Course Theory
Book 2. Olympiad Inequalities
Book 2. Olympiad Inequalities
- 1. Basic Inequality Principles
- 2. AM-GM
- 3. Cauchy-Schwarz
- 4. Rearrangement and Chebyshev
- 5. Jensen's Inequality Intro
- 6. UVW Method and Symmetric Inequalities
- 7. Homogeneous Inequalities
- 8. Substitution Methods
- 9. Inequalities with Constraints
- 10. Hard Inequality Problems
- 11. Mixed Inequality Sets
- 12. Mock Olympiads: Inequalities
Chapter
Basic Inequality Principles
Key idea
The basic olympiad principle for inequalities is this: before estimating an expression, understand where equality could occur and which quantity is the right one to compare. Many solutions use no heavy theorem; the decisive step is choosing a square, an ordering, or an extremal element.
Basic facts
We shall use \(t^2\ge0\), \((x-y)^2\ge0\), ordering of numbers, the minimal sum of distinct positive integers, sign checks before multiplying inequalities, and equality-case analysis. Remember: an inequality may be multiplied by an expression only after its sign is controlled.
When to use this method
Use these principles when the problem contains words such as “largest”, “smallest”, “distinct positive”, “fixed sum”, “prove impossible”, “find the maximum”, or “equality occurs only when...”. These are signals to look for a boundary, an invariant, or a square.
How to recognise the method
If the expression is symmetric, try to write it as a sum of squares of differences. If the numbers are ordered, compare the extreme elements. If the condition holds for every pair or every partition, look for a global estimate: a sum, a product, or the number of positive/negative elements.
Typical mistakes
Do not multiply inequalities before checking signs. Do not forget equality cases. Another common mistake is estimating each term separately when the problem actually requires an extremal element or the whole set at once.
Mini-checklist
1. Where can equality occur? 2. Which expressions are certainly nonnegative? 3. Can the objects be ordered? 4. Is multiplying inequalities safe? 5. Is there an extremal element? 6. Are all impossible cases really excluded?
Example 1. Square of a difference
What this example teaches. A basic estimate is often hidden in a square.
Problem. Prove that \(a^2+b^2\ge2ab\).
Move everything to one side: \(a^2-2ab+b^2=(a-b)^2\ge0\). Hence \(a^2+b^2\ge2ab\). Equality occurs when \(a=b\).
Example 2. Equality case
What this example teaches. In an inequality, one must see both the estimate and when it is sharp.
Problem. If \(x+y=10\), find the minimum value of \(x^2+y^2\).
\(x^2+y^2=\frac{(x+y)^2+(x-y)^2}{2}=50+\frac{(x-y)^2}{2}\ge50\). Equality occurs when \(x=y=5\).
Example 3. Estimating a fraction
What this example teaches. The maximum of a fraction can sometimes be found without calculus.
Problem. Prove that for \(x>0\), \(\frac{x}{x^2+x+1}\le\frac13\).
The denominator is positive, so the inequality is equivalent to \(3x\le x^2+x+1\), i.e. \((x-1)^2\ge0\).
Example 4. Ordering
What this example teaches. If extreme elements are present, use them.
Problem. Let \(0 Since the numbers are distinct positive integers, \(a_i\ge i\). Therefore the sum is at least \(1+2+\cdots+n=\frac{n(n+1)}2\).
Example 5. Careful multiplication
What this example teaches. Before multiplying inequalities, signs must be known.
Problem. If \(u>v>0\) and \(p>q>0\), prove \(up+vq>uq+vp\).
The difference is \(up+vq-uq-vp=(u-v)(p-q)>0\). The signs of both factors are controlled, so the conclusion is rigorous.
Example 6. Inequality as impossibility
What this example teaches. Sometimes one must prove that equality cannot occur in several places at once.
Problem. Can distinct real numbers \(x,y,z\) satisfy \(\frac1{x^2+x+1}+\frac1{y^2+y+1}+\frac1{z^2+z+1}=4\)?
\(x^2+x+1=\left(x+\frac12\right)^2+\frac34\), so each term is at most \(\frac43\). The sum is at most \(4\), and equality would require \(x=y=z=-\frac12\), contradicting distinctness.
Example 7. Extremal element
What this example teaches. The largest or smallest element often cannot satisfy the same rule as the others.
Problem. Positive numbers are written around a circle. Can the largest number be equal to the absolute difference of its two neighbours?
No. Both neighbours are positive and do not exceed the largest number \(M\), so the absolute difference of the neighbours is strictly smaller than \(M\).
Example 8. Invariant
What this example teaches. In operation problems, it is often useful to count types of objects rather than exact objects.
Problem. A rectangle has both sides greater than \(1\). An operation either replaces it by the rectangle with reciprocal sides, or splits one side into two positive factors. Why is it impossible to end with only rectangles whose one side is greater than \(1\) and the other is less than \(1\)?
Call a rectangle uniform if both sides are greater than \(1\) or both sides are less than \(1\). The first operation preserves the number of uniform rectangles, and the second operation applied to a uniform rectangle leaves at least one uniform rectangle. We started with one uniform rectangle, so they cannot all disappear.
Chapter
AM-GM
Key idea
AM-GM is useful not merely because positive numbers appear, but because one can choose terms whose product is controlled. The olympiad difficulty is not applying the formula, but preparing the expression for it.
Basic facts
For \(x_1,\ldots,x_n>0\), \(\frac{x_1+\cdots+x_n}{n}\ge\sqrt[n]{x_1\cdots x_n}\). Weighted form: if the weights are positive and sum to \(1\), then \(\lambda_1x_1+\cdots+\lambda_nx_n\ge x_1^{\lambda_1}\cdots x_n^{\lambda_n}\). Equality occurs when all terms chosen for AM-GM are equal.
When to use this method
Use it for fixed sums, fixed products, expressions like \(x+\frac{k}{x}\), products of several positive quantities, and maximum/minimum problems with an equality case.
How to recognise the method
Look for “positive numbers”, “sum equals”, “product equals”, “find the greatest/smallest”. If the expression does not fit immediately, try multiplying, dividing by a fixed quantity, ordering the variables, or substituting part of the condition.
Typical mistakes
Do not apply AM-GM to nonpositive quantities. Do not forget that equality requires equality of the exact terms to which AM-GM was applied. A common mistake is finding a correct bound without checking whether equality is compatible with the conditions.
Mini-checklist
1. Are all chosen terms positive? 2. Is their product known or bounded? 3. Is equality compatible with the condition? 4. Is a weighted form needed? 5. Is normalization or ordering needed first?
Example 1. Two terms
What this example teaches. The basic AM-GM form.
Problem. For \(x>0\), prove \(x+\frac{25}{x}\ge10\).
By AM-GM, \(x+\frac{25}{x}\ge2\sqrt{25}=10\). Equality occurs at \(x=5\).
Example 2. Fixed product
What this example teaches. Minimum of a sum when the product is known.
Problem. If \(abc=8\), prove \(a+b+c\ge6\).
\(\frac{a+b+c}{3}\ge\sqrt[3]{abc}=2\), hence \(a+b+c\ge6\). Equality occurs when \(a=b=c=2\).
Example 3. Fixed sum
What this example teaches. Maximum of a product when the sum is known.
Problem. If \(a+b+c=9\), prove \(abc\le27\).
\(\sqrt[3]{abc}\le\frac{a+b+c}{3}=3\), so \(abc\le27\). Equality occurs when \(a=b=c=3\).
Example 4. Weighted idea
What this example teaches. Sometimes one variable must be used several times.
Problem. If \(x^2y=1\), prove \(2x+y\ge3\).
Apply AM-GM to \(x,x,y\): \(\frac{x+x+y}{3}\ge\sqrt[3]{x^2y}=1\). Therefore \(2x+y\ge3\).
Example 5. Preparing the expression
What this example teaches. Before AM-GM, one may rewrite the condition.
Problem. If \(ab+bc+ca=1\), prove \(\sqrt{a+\frac1a}\ge\sqrt b+\sqrt c\).
\(a+\frac1a=a+\frac{ab+ac+bc}{a}=a+b+c+\frac{bc}{a}\). By AM-GM, \(a+\frac{bc}{a}\ge2\sqrt{bc}\). Hence \(a+\frac1a\ge(\sqrt b+\sqrt c)^2\).
Example 6. Equality in several places
What this example teaches. Equality cases must be compatible with each other.
Problem. If \(a,b,c>0\), \(a+b+c=3\), prove \((1+a)(1+b)(1+c)\le8\).
The sum of \(1+a,1+b,1+c\) is \(6\). By AM-GM, their product is at most \(\left(\frac63\right)^3=8\). Equality occurs when \(a=b=c=1\).
Example 7. AM-GM after ordering
What this example teaches. In harder problems, one first chooses an order of variables.
Problem. If \(a\ge b\ge c\ge d>0\) and \(a+b+c+d=3\), prove \(ab(c+d)\le1\).
By AM-GM for \(a,b,c+d\): \(ab(c+d)\le\left(\frac{a+b+c+d}{3}\right)^3=1\).
Example 8. Product of fractions
What this example teaches. First prove a product estimate, then apply AM-GM to the sum.
Problem. Prove for \(x_i>0\): \(\sum_{i=1}^{n}\frac{1+x_i^2}{1+x_ix_{i+1}}\ge n\), where \(x_{n+1}=x_1\).
For every \(i\), \((1+x_i^2)(1+x_{i+1}^2)\ge(1+x_ix_{i+1})^2\). Multiplying gives that the product of the fractions is at least \(1\). Then their sum is at least \(n\) by AM-GM.
Chapter
Cauchy-Schwarz
Key idea
Cauchy-Schwarz can replace a sum of fractions by one global fraction. In olympiad problems, the main task is choosing numerators and denominators so that the denominator sum becomes simple.
Basic facts
Standard form: \((a_1^2+\cdots+a_n^2)(b_1^2+\cdots+b_n^2)\ge(a_1b_1+\cdots+a_nb_n)^2\). Engel form: for \(q_i>0\), \[\frac{x_1^2}{q_1}+\cdots+\frac{x_n^2}{q_n}\ge\frac{(x_1+\cdots+x_n)^2}{q_1+\cdots+q_n}.\]
When to use this method
Use it for sums such as \(\frac{x^2}{A}+\frac{y^2}{B}+\frac{z^2}{C}\), fractions with cyclic denominators, reciprocal estimates, and problems where the sum of denominators is controlled by the condition.
How to recognise the method
If a numerator is a square or can be made into a square, try Engel form. If the numerator is not a square, rewrite \(\frac{a}{b}\) as \(\frac{(\sqrt{a})^2}{b}\), or multiply the fraction into a useful form.
Typical mistakes
Do not forget positivity of denominators. A common mistake is applying Cauchy with the wrong numerators and getting a useless denominator sum. Another mistake is losing the equality case.
Mini-checklist
1. Are all denominators positive? 2. Which numerators give the needed sum? 3. Does the denominator sum simplify? 4. Is a preliminary denominator estimate needed? 5. Is equality compatible with the condition?
Example 1. Standard form
Problem. Prove \((a^2+b^2)(c^2+d^2)\ge(ac+bd)^2\).
This is Cauchy for vectors \((a,b)\) and \((c,d)\). After expansion, the difference is \((ad-bc)^2\ge0\).
Example 2. Engel form
Problem. Prove \(\frac{x^2}{a}+\frac{y^2}{b}\ge\frac{(x+y)^2}{a+b}\) for \(a,b>0\).
This is Engel form: \(\sum\frac{x_i^2}{q_i}\ge\frac{(\sum x_i)^2}{\sum q_i}\).
Example 3. Nesbitt
Problem. For \(a,b,c>0\), prove \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac32\).
Write \(a=\frac{a^2}{a}\): \(\sum\frac{a}{b+c}=\sum\frac{a^2}{a(b+c)}\ge\frac{(a+b+c)^2}{2(ab+bc+ca)}\ge\frac32\).
Example 4. Denominator sum
Problem. Prove \(\sum\frac{x^2}{x+y}\ge\frac{x+y+z}{2}\).
By Cauchy, \(\sum\frac{x^2}{x+y}\ge\frac{(x+y+z)^2}{2(x+y+z)}=\frac{x+y+z}{2}\).
Example 5. Reciprocals
Problem. If \(a+b+c=1\), prove \(\frac1a+\frac1b+\frac1c\ge9\).
By Cauchy, \((1+1+1)^2\le(a+b+c)\left(\frac1a+\frac1b+\frac1c\right)\). Since \(a+b+c=1\), the result follows.
Example 6. Creating a square
Problem. Prove \(\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}\ge\frac{(\sqrt{x}+\sqrt{y}+\sqrt{z})^2}{2(x+y+z)}\).
Write the numerators as \((\sqrt{x})^2\), \((\sqrt{y})^2\), \((\sqrt{z})^2\), then apply Cauchy.
Example 7. Preliminary estimate
Problem. For \(t,u>0\), estimate \(\frac{t^3}{t^2+u}\) from below using \(t\) and \(\sqrt{u}\).
\(\frac{t^3}{t^2+u}=t-\frac{tu}{t^2+u}\ge t-\frac{tu}{2t\sqrt{u}}=t-\frac{\sqrt{u}}{2}\).
Example 8. Mixed step
Problem. If \(a+b+c=3\), prove \(ab+bc+ca\le3\).
\((a+b+c)^2\ge3(ab+bc+ca)\), since this is equivalent to \(\frac12((a-b)^2+(b-c)^2+(c-a)^2)\ge0\).
Chapter
Rearrangement and Chebyshev
Key idea
If two sequences are sorted in the same order, pairing them term by term gives the largest possible product sum. If they are sorted in opposite orders, the product sum is smallest. Chebyshev's inequality turns this ordering idea into an estimate for the average product.
Basic facts
Rearrangement inequality. If \(a_1\le\cdots\le a_n\) and \(b_1\le\cdots\le b_n\), then for any permutation \(\sigma\)
\[\sum_{i=1}^n a_i b_i\ge \sum_{i=1}^n a_i b_{\sigma(i)}\ge \sum_{i=1}^n a_i b_{n+1-i}.\]
Chebyshev's inequality. For similarly ordered sequences,
\[\frac1n\sum_{i=1}^n a_i b_i\ge \left(\frac1n\sum_{i=1}^n a_i\right)\left(\frac1n\sum_{i=1}^n b_i\right).\]
When to use this method
Use the method when a problem involves ordering, permutations, a maximum or minimum product sum, or power sums such as \(\sum a_i^{m+1}\), which can be viewed as products \(a_i^m\cdot a_i\).
How to recognise the method
Look for two monotone sequences: \(a_i\) and \(a_i^2\), \(a_i^2\) and \(a_i^3\), numbers and their reciprocals in the opposite order, or coefficients and variables with zero sum. If the expression changes when letters are permuted, rearrangement is often worth testing.
Typical mistakes
Do not apply the inequality before checking the order. Reciprocals have the opposite order. In symmetric problems one may rename variables first, but in cyclic expressions this must be done carefully: after renaming, the cyclic sum is only some permutation of products.
Mini-checklist
1. Which two sequences should be ordered? 2. Are they similarly or oppositely ordered? 3. Is the goal a maximum, a minimum, or an average estimate? 4. Can the cyclic sum be treated as an arbitrary permutation? 5. When can equality occur?
Example 1. One swap
The first step of the method is understanding why the correct pairing is better than the incorrect one.
Problem. Let \(a\le b\) and \(x\le y\). Prove that \(ax+by\ge ay+bx\).
Subtract the right side from the left side:
\[ax+by-ay-bx=(b-a)(y-x)\ge0.\]
One adjacent swap cannot increase the sum once the order is already correct.
Example 2. Maximum permutation
Any permutation can be repaired by a sequence of adjacent swaps.
Problem. If \(a\le b\le c\) and \(x\le y\le z\), show that the maximum of \(ap+bq+cr\), where \(p,q,r\) is a permutation of \(x,y,z\), is \(ax+by+cz\).
If a larger coefficient is paired with a smaller variable and a smaller coefficient with a larger variable, swapping them increases the sum by \((b-a)(y-x)\ge0\). Repeating such swaps gives the same order. Hence the maximum is \(ax+by+cz\).
Example 3. Chebyshev's inequality
Chebyshev is the average form of rearrangement.
Problem. Let \(a\le b\le c\) and \(x\le y\le z\). Prove \[3(ax+by+cz)\ge(a+b+c)(x+y+z).\]
Expanding the difference gives
\[3\sum ax-\sum a\sum x=(b-a)(y-x)+(c-a)(z-x)+(c-b)(z-y)\ge0.\]
Example 4. Opposite order
Reciprocals reverse the order.
Example 5. Power sums
Powers automatically preserve the order.
Problem. Prove for \(x,y,z\ge0\): \[3(x^3+y^3+z^3)\ge(x+y+z)(x^2+y^2+z^2).\]
Reorder the numbers so that \(x\le y\le z\). Then \(x^2\le y^2\le z^2\). By Chebyshev,
\[\frac{x^3+y^3+z^3}{3}\ge\frac{x+y+z}{3}\cdot\frac{x^2+y^2+z^2}{3}.\]
Example 6. Cyclic sum
A cyclic sum is often just one of the possible permutations.
Problem. Prove for \(a,b,c>0\): \[a^4+b^4+c^4\ge a^3b+b^3c+c^3a.\]
Order \(a,b,c\) as \(u_1\le u_2\le u_3\). Then \(u_1^3\le u_2^3\le u_3^3\). The largest product sum \(u_i^3\cdot u_{\sigma(i)}\) is \(\sum u_i^4\). The right side is one of these permutations, so it is not larger than the left side.
Example 7. Zero sum
Chebyshev often turns a zero-sum condition into the sign of a dot product.
Problem. Let \(a\le b\le c\), \(x\le y\le z\), and \(x+y+z=0\). Prove that \(ax+by+cz\ge0\) if \(a+b+c\ge0\).
By Chebyshev, \(3(ax+by+cz)\ge(a+b+c)(x+y+z)=0\). Hence \(ax+by+cz\ge0\).
Example 8. Keeping equality cases
In permutation problems, it is important to know when every swap gives zero gain.
Problem. For \(a\le b\le c\) and \(x\le y\le z\), find when \(ax+by+cz=az+by+cx\).
The difference is \((c-a)(z-x)\). Therefore equality holds if and only if \(a=c\) or \(x=z\), meaning one of the sequences is constant at the extreme positions.
Chapter
Jensen's Inequality Intro
Key idea
Jensen's inequality says that for a convex function, the value at the average is not greater than the average value of the function; for a concave function, the direction reverses. In olympiad problems, the real skill is not naming convexity, but choosing the right function for the given constraint.
Basic facts
If \(f\) is convex, then
\[\frac{f(x_1)+\cdots+f(x_n)}{n}\ge f\left(\frac{x_1+\cdots+x_n}{n}\right).\]
If \(f\) is concave, the inequality reverses. Common functions: \(x^2\), \(x^4\), \(1/x\), \(e^x\) are convex on the relevant domains; \(\sqrt{x}\) and \(\ln x\) are concave for \(x>0\). A tangent to a convex function lies below the graph, while a tangent to a concave function lies above the graph.
When to use this method
Use the method with a fixed sum, an average value, products after taking logarithms, sums of roots, reciprocals, and problems where equality is expected at \(x_1=\cdots=x_n\).
How to recognise the method
Look for expressions of the form \(\sum f(x_i)\) and a condition on \(\sum x_i\). If a product must be estimated, take logarithms. If \(x^x\) appears, consider \(x\ln x\). If the problem asks for the maximum of a sum of roots or logarithms, concavity is likely involved.
Typical mistakes
Do not forget domains: \(\ln x\), \(\sqrt{x}\), and \(1/x\) require positivity or nonnegativity. A common mistake is reversing the inequality in the wrong direction for a concave function. Another mistake is applying Jensen when no fixed average is available.
Mini-checklist
1. Which function \(f\) is hidden in the expression? 2. Is it convex or concave on the required domain? 3. What is known about the average of the arguments? 4. Where should equality occur? 5. Is a logarithm or tangent line needed first?
Example 1. Convexity of the square
This example shows the basic direction of Jensen for a convex function.
Problem. Prove \(a^2+b^2+c^2\ge\frac{(a+b+c)^2}{3}\).
The function \(f(x)=x^2\) is convex. Hence \(\frac{a^2+b^2+c^2}{3}\ge\left(\frac{a+b+c}{3}\right)^2\). Multiply by \(3\).
Example 2. Concavity of the square root
For concave functions, the inequality direction reverses.
Problem. For \(x,y,z\ge0\), \(x+y+z=12\), prove \(\sqrt{x}+\sqrt{y}+\sqrt{z}\le6\).
The function \(\sqrt{x}\) is concave. Therefore \(\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{3}\le\sqrt{\frac{x+y+z}{3}}=\sqrt{4}=2\).
Example 3. Reciprocals
The convexity of \(1/x\) gives a classical lower bound.
Problem. Prove for \(a,b,c>0\): \(\frac1a+\frac1b+\frac1c\ge\frac{9}{a+b+c}\).
For \(f(x)=1/x\), the function is convex on \(x>0\). By Jensen, \(\frac13\left(\frac1a+\frac1b+\frac1c\right)\ge\frac{1}{(a+b+c)/3}\). This gives the result.
Example 4. Logarithm and product
Products are often easier to estimate after taking logarithms.
Problem. If \(a,b,c>0\) and \(a+b+c=3\), prove \(abc\le1\).
The function \(\ln x\) is concave. Hence \(\frac{\ln a+\ln b+\ln c}{3}\le\ln\frac{a+b+c}{3}=\ln1=0\). Thus \(\ln(abc)\le0\), so \(abc\le1\).
Example 5. Tangent to the square
A tangent line often replaces a function by a linear bound.
Problem. Prove \(x^2\ge2x-1\) for all real \(x\).
This is the tangent to \(x^2\) at \(1\). Algebraically, \(x^2-2x+1=(x-1)^2\ge0\).
Example 6. Tangent to the logarithm
For a concave function, the tangent lies above the graph.
Problem. Prove \(\ln x\le x-1\) for \(x>0\).
The function \(\ln x\) is concave, so its graph lies below the tangent at \(1\). The tangent is \(y=x-1\). Hence \(\ln x\le x-1\).
Example 7. Weighted Jensen
Sometimes the average has unequal weights.
Problem. Prove for \(x,y\ge0\): \(2\sqrt{x}+\sqrt{y}\le3\sqrt{\frac{2x+y}{3}}\).
Use the concavity of \(\sqrt{x}\) with weights \(\frac23\) and \(\frac13\): \(\frac23\sqrt{x}+\frac13\sqrt{y}\le\sqrt{\frac{2x+y}{3}}\). Multiply by \(3\).
Example 8. \(x\ln x\)
Expressions \(x^x\) often reduce to the function \(x\ln x\).
Problem. Let \(x,y,z>0\), \(x+y+z=3\). Prove \(x^x y^y z^z\ge1\).
The function \(f(t)=t\ln t\) is convex for \(t>0\). By Jensen, \(\frac{x\ln x+y\ln y+z\ln z}{3}\ge 1\cdot\ln1=0\). Hence \(\ln(x^x y^y z^z)\ge0\), so \(x^x y^y z^z\ge1\).
Chapter
UVW Method and Symmetric Inequalities
Key idea
The UVW method works with symmetric inequalities in three variables. Instead of \(a,b,c\), we look at \(p=a+b+c\), \(q=ab+bc+ca\), \(r=abc\). Often, after fixing \(p,q\), the expression depends on \(r\) in a simple way, and the hardest case occurs on the boundary or when two variables are equal.
Basic facts
Main notation: \(p=a+b+c\), \(q=ab+bc+ca\), \(r=abc\). Useful formulas:
\[a^2+b^2+c^2=p^2-2q,\qquad a^3+b^3+c^3=p^3-3pq+3r.\]
Also \(\sum_{\mathrm{sym}}a^2b=pq-3r\), where \(\sum_{\mathrm{sym}}a^2b=a^2b+a^2c+b^2a+b^2c+c^2a+c^2b\). Schur degree \(3\): \[p^3-4pq+9r\ge0.\]
When to use this method
Use the method when the inequality is symmetric in \(a,b,c\), equality is expected at \(a=b=c\) or at two equal variables, and the expression can be rewritten through \(p,q,r\). It is especially useful for degrees \(3\) and \(4\), Schur-type inequalities, and fixed-sum problems.
How to recognise the method
Check whether the problem is unchanged by permuting \(a,b,c\). Then identify the degree and try to express the sums through \(p,q,r\). If after fixing \(p,q\) the dependence on \(r\) is linear, it is natural to check the cases \(b=c\) and \(c=0\).
Typical mistakes
Do not apply UVW to a nonsymmetric expression. Do not confuse the symmetric sum \(\sum_{\mathrm{sym}}\) with a cyclic sum. Another mistake is checking only \(a=b=c\): a proof usually requires the two-equal case or the boundary.
Mini-checklist
1. Is the inequality symmetric? 2. Which \(p,q,r\) appear? 3. Are \(p\) or \(q\) fixed? 4. Can Schur be applied? 5. What do the cases \(b=c\) and \(c=0\) give? 6. Does the equality case match the answer?
Example 1. Translation to \(p,q,r\)
Problem. Express \(a^2+b^2+c^2\) and \(a^3+b^3+c^3\) through \(p,q,r\).
From \(p^2=a^2+b^2+c^2+2q\), we get \(\sum a^2=p^2-2q\). Also \(p^3=\sum a^3+3\sum_{\mathrm{sym}}a^2b+6r\), and \(\sum_{\mathrm{sym}}a^2b=pq-3r\), so \(\sum a^3=p^3-3pq+3r\).
Example 2. Symmetric sum
Problem. Prove \(\sum_{\mathrm{sym}}a^2b=pq-3r\).
Expand \(pq=(a+b+c)(ab+bc+ca)\). We get all six terms \(a^2b,a^2c,\ldots,c^2b\), plus \(3abc\). Hence \(\sum_{\mathrm{sym}}a^2b=pq-3r\).
Example 3. Schur degree 3
Problem. Prove \(\sum a^3+3abc\ge\sum_{\mathrm{sym}}a^2b\).
The left side minus the right side equals \((p^3-3pq+3r)+3r-(pq-3r)=p^3-4pq+9r\). This is Schur degree \(3\), hence it is nonnegative.
Example 4. Checking two equal variables
Problem. Check Schur degree \(3\) in the case \(b=c=1\), \(a=t\).
The inequality becomes \(t^3+2+3t\ge 2t^2+2t+2\), or \(t(t-1)^2\ge0\). This is a typical two-equal-variable check.
Example 5. Maximum of \(q\)
Problem. Prove \(p^2\ge3q\).
\(p^2-3q=a^2+b^2+c^2-ab-bc-ca=\frac12((a-b)^2+(b-c)^2+(c-a)^2)\ge0\).
Example 6. Maximum of \(r\)
Problem. If \(a+b+c=p\), prove \(abc\le\frac{p^3}{27}\).
By AM-GM, \(\frac{a+b+c}{3}\ge\sqrt[3]{abc}\). Therefore \(abc\le(p/3)^3\). In UVW language, this agrees with the fact that for fixed \(p\), the maximum of the symmetric product occurs at \(a=b=c\).
Example 7. Schur degree 4
Problem. Prove \(\sum a^4+abc(a+b+c)\ge\sum_{\mathrm{sym}}a^3b\).
This is Schur degree \(4\). By UVW, it is enough to check the two-equal case and the boundary. For \(b=c=1\), \(a=t\), the difference is \(t^2(t-1)^2\ge0\); on the boundary we get an obvious sum of squares. Hence the inequality holds.
Example 8. Not every cyclic expression is symmetric
Problem. Why can UVW not be applied directly to \(a^3b+b^3c+c^3a\le a^4+b^4+c^4\)?
The right side is symmetric, but the left side is cyclic and changes under permutations of \(a,b,c\). Therefore UVW does not apply directly; this problem is better handled by rearrangement or by summing over all permutations.
Chapter
Homogeneous Inequalities
Key idea
A homogeneous inequality keeps its essential form when all variables are multiplied by the same positive number. Therefore we may choose a convenient normalization: \(a+b+c=1\), \(abc=1\), \(a+b+c=3\), or \(c=1\).
Basic facts
An expression has degree \(d\) if replacing \(a,b,c\) by \(ta,tb,tc\) multiplies it by \(t^d\). An inequality is homogeneous if both sides have the same degree. For example, \(\sum a^2\ge ab+bc+ca\) has degree \(2\), while \(\sum a/(b+c)\ge3/2\) has degree \(0\).
When to use this method
Normalization is useful when the expression is bulky but all terms have the same degree. Fix the sum for problems involving \(a+b+c\), fix the product for problems involving \(abc\), and fix one variable for two-variable or degree-zero fraction problems.
How to recognise the method
Check what happens under \(a,b,c\to ta,tb,tc\). If both sides are multiplied by the same factor, normalization is allowed. If a constant appears, make sure the expression has degree \(0\), or first homogenize the constant using \(a+b+c\), \(abc\), or another fixed quantity.
Typical mistakes
Do not set \(a+b+c=1\) in a nonhomogeneous inequality. Do not set \(abc=1\) if variables may be zero. Another mistake is forgetting to return to the original form after normalization or losing the equality case.
Mini-checklist
1. What is the degree of each side? 2. Are all variables positive? 3. Which normalization simplifies the expression? 4. What happens to equality? 5. Must a constant be homogenized first?
Example 1. Checking degree
Problem. Find the degree of \(\frac{a^2}{b+c}\), \(\frac{a}{b+c}\), \(abc(a+b+c)\).
The first has degree \(1\), the second degree \(0\), and the third degree \(4\). This is seen after replacing \(a,b,c\) by \(ta,tb,tc\).
Example 2. Sum normalization
Problem. Prove \(\sum a^2\ge\frac{(a+b+c)^2}{3}\).
The inequality is homogeneous of degree \(2\). We may set \(a+b+c=1\). Then we need \(\sum a^2\ge1/3\), which follows from Cauchy: \((a+b+c)^2\le3\sum a^2\).
Example 3. Product normalization
Problem. Prove \(a+b+c\ge3\sqrt[3]{abc}\).
If \(abc>0\), divide the variables by \(\sqrt[3]{abc}\) and obtain a new triple with product \(1\). Then the problem becomes \(x+y+z\ge3\), which is AM-GM.
Example 4. Degree zero
Problem. Prove \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac32\).
The expression has degree \(0\), so we may set \(a+b+c=1\). Then \(b+c=1-a\), and the inequality becomes \(\sum\frac{a}{1-a}\ge3/2\), which follows from Nesbitt or Cauchy.
Example 5. Homogenizing a constant
Problem. If \(a+b+c=1\), prove \(ab+bc+ca\le\frac13\).
In homogeneous form, this is \(ab+bc+ca\le\frac{(a+b+c)^2}{3}\), or \(3q\le p^2\). This is true because \(p^2-3q=\frac12\sum(a-b)^2\ge0\).
Example 6. Normalizing one variable
Problem. Why may we set \(b=1\) in the inequality \(\frac{a^2+b^2}{ab}\ge2\)?
The expression has degree \(0\): replacing \(a,b\) by \(ta,tb\) does not change it. Thus we may divide both variables by \(b\) and get \(x=a/b\), \(b=1\).
Example 7. Degree-one fractions
Problem. Prove \(\sum\frac{a^2}{b+c}\ge\frac{a+b+c}{2}\).
By Cauchy, \(\sum\frac{a^2}{b+c}\ge\frac{(a+b+c)^2}{2(a+b+c)}=\frac{a+b+c}{2}\). Homogeneity explains why we could have set \(a+b+c=1\) first.
Example 8. Checking whether normalization is allowed
Problem. Why can we not set \(a+b+c=1\) in \(a^2+b^2+c^2\ge1\) without an extra condition?
The left side has degree \(2\), while the right side has degree \(0\). The inequality is not homogeneous and changes under scaling. Such normalization changes the problem.
Chapter
Substitution Methods
Key idea
A substitution is not decorative; it turns an expression into a more familiar form: removing ratios, encoding \(abc=1\), replacing triangle sides by sums, removing roots, or using the equality case.
Basic facts
For ratios, one often sets \(x=a/b\). If \(abc=1\), look for variables with product \(1\), or write \(a=x/y\), \(b=y/z\), \(c=z/x\). If \(a,b,c\) are triangle sides, use \(a=y+z\), \(b=z+x\), \(c=x+y\), where \(x,y,z>0\).
When to use this method
Use substitution when the condition is awkward, fractions contain repeated ratios, roots \(\sqrt{a}\) appear, triangle sides are involved, a constraint \(x^2+y^2=1\) appears, or equality is expected at \(a=b=c\).
How to recognise the method
If \(a/b,b/c,c/a\) appear, look for product \(1\). If \(b+c-a\), \(c+a-b\), \(a+b-c\) appear, the triangle substitution is probably needed. If the expression contains \(\sqrt{ab}\), try \(a=x^2\), \(b=y^2\).
Typical mistakes
Do not substitute without checking reversibility and domain: \(x,y,z\) must be positive where required. In the triangle substitution, remember \(x=(b+c-a)/2\), \(y=(c+a-b)/2\), \(z=(a+b-c)/2\). After solving, return to the original variables.
Mini-checklist
1. Which condition is awkward? 2. What becomes simpler after substitution? 3. Is the substitution reversible? 4. Is positivity preserved? 5. Where is the equality case?
Example 1. The ratio \(a/b\)
Problem. Prove \(\frac{a}{b}+\frac{b}{a}\ge2\).
Set \(x=a/b>0\). Then we need \(x+\frac1x\ge2\), which is equivalent to \((x-1)^2\ge0\).
Example 2. Cyclic ratios
Problem. Prove \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge3\).
Set \(x=a/b\), \(y=b/c\), \(z=c/a\). Then \(xyz=1\), and by AM-GM \(x+y+z\ge3\sqrt[3]{xyz}=3\).
Example 3. Triangle substitution
Problem. If \(a,b,c\) are triangle sides, explain the substitution \(a=y+z\), \(b=z+x\), \(c=x+y\).
Take \(x=(b+c-a)/2\), \(y=(c+a-b)/2\), \(z=(a+b-c)/2\). For triangle sides these numbers are positive, and then \(a=y+z\), \(b=z+x\), \(c=x+y\).
Example 4. Triangle sides
Problem. For the sides of a triangle, prove \(a^2+b^2+c^2<2(ab+bc+ca)\).
Let \(a=y+z\), \(b=z+x\), \(c=x+y\). Then \(2(ab+bc+ca)-(a^2+b^2+c^2)=4(xy+yz+zx)>0\).
Example 5. Removing roots
Problem. Prove \(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le a+b+c\).
Set \(a=x^2\), \(b=y^2\), \(c=z^2\). Then we need \(xy+yz+zx\le x^2+y^2+z^2\), which follows from \(\frac12((x-y)^2+(y-z)^2+(z-x)^2)\ge0\).
Example 6. Deviations from equality
Problem. If \(a+b+c=3\), prove \(\sum a^2\ge3\).
Set \(a=1+x\), \(b=1+y\), \(c=1+z\), where \(x+y+z=0\). Then \(\sum a^2=3+2(x+y+z)+x^2+y^2+z^2=3+\sum x^2\ge3\).
Example 7. Trigonometric preview
Problem. If \(x^2+y^2=1\), \(x,y\ge0\), prove \(xy\le1/2\).
We may write \(x=\cos t\), \(y=\sin t\). Then \(xy=\frac12\sin 2t\le1/2\). This is only a preview: algebra is often enough.
Example 8. Substitution from denominators
Problem. For triangle sides, prove \(\sum\frac{a}{b+c-a}\ge3\).
Let \(a=y+z\), \(b=z+x\), \(c=x+y\). Then \(b+c-a=2x\), and the sum equals \(\frac12\sum\frac{y+z}{x}\). Since \(\sum\frac{y}{x}+\frac{x}{y}\ge6\), the original sum is at least \(3\).
Chapter
Inequalities with Constraints
Key idea
A constraint in an inequality usually tells us where to look for the extremum. With a fixed sum, expressions involving convex functions are often minimized at equal variables, while expressions involving concave functions are often maximized at equal variables. With a fixed product, AM-GM and the normalization \(abc=1\) are natural.
Basic facts
If \(a+b+c=S\), then \(\sum a^2\ge S^2/3\), \(ab+bc+ca\le S^2/3\), and \(abc\le(S/3)^3\). If \(abc=P\), then \(a+b+c\ge3\sqrt[3]{P}\) and \(ab+bc+ca\ge3\sqrt[3]{P^2}\). With fixed \(q=ab+bc+ca\), the relations \(p^2\ge3q\) and \(q^2\ge3pr\) are useful.
When to use this method
Use this approach when the condition fixes a sum, product, \(ab+bc+ca\), sum of squares, or average value. The problem often becomes an extremum problem for one symmetric quantity.
How to recognise the method
If equality is expected at \(a=b=c\), first try Jensen, AM-GM, Cauchy, or UVW. If variables are nonnegative and the sum is fixed, the maximum of some expressions may occur on the boundary: for example, \(\sum a^2\) is maximal at \(S,0,0\).
Typical mistakes
Do not confuse maximum and minimum: with a fixed sum, \(\sum a^2\) is minimized at equality but maximized on the boundary. Do not use \(abc=1\) if variables may be zero. Always check whether equality is attainable in the allowed domain.
Mini-checklist
1. What is fixed? 2. Are we looking for a maximum or a minimum? 3. Is equality at equal variables or on the boundary? 4. Is Jensen, AM-GM, Cauchy, or UVW needed? 5. Has the equality case been checked?
Example 1. Minimum of squares
With a fixed sum, the sum of squares is minimized at equality.
Problem. Let \(a+b+c=6\). Find the minimum of \(a^2+b^2+c^2\).
By Cauchy, \((a+b+c)^2\le3(a^2+b^2+c^2)\). Hence \(36\le3\sum a^2\), so \(\sum a^2\ge12\). Equality holds at \(a=b=c=2\).
Example 2. Maximum product
With a fixed sum, the product of positive numbers is maximized at equality.
Problem. If \(a+b+c=9\), find the maximum of \(abc\).
By AM-GM, \(\sqrt[3]{abc}\le\frac{a+b+c}{3}=3\). Therefore \(abc\le27\). Equality holds at \(a=b=c=3\).
Example 3. Fixed product
With a fixed product, the sum is minimized at equality.
Problem. If \(abc=8\), prove \(a+b+c\ge6\).
By AM-GM, \(a+b+c\ge3\sqrt[3]{abc}=6\). Equality holds at \(a=b=c=2\).
Example 4. Fixed sum and \(q\)
The quantity \(q=ab+bc+ca\) is controlled by the sum.
Problem. If \(a+b+c=1\), prove \(ab+bc+ca\le1/3\).
\((a+b+c)^2\ge3(ab+bc+ca)\). With \(a+b+c=1\), we get \(q\le1/3\). Equality holds at \(a=b=c=1/3\).
Example 5. Sum of roots
The concavity of the square root gives the maximum at equal variables.
Problem. Let \(x+y+z=12\), \(x,y,z\ge0\). Find the maximum of \(\sqrt{x}+\sqrt{y}+\sqrt{z}\).
By Jensen for \(\sqrt{x}\), \(\frac{\sum\sqrt{x}}{3}\le\sqrt{\frac{x+y+z}{3}}=2\). The maximum is \(6\), attained at \(x=y=z=4\).
Example 6. Reciprocals
A fixed sum and convexity of \(1/x\) give the minimum of the sum of reciprocals.
Problem. If \(a+b+c=3\), prove \(\frac1a+\frac1b+\frac1c\ge3\).
By Cauchy, \((1+1+1)^2\le(a+b+c)(1/a+1/b+1/c)\). Since the sum is \(3\), the reciprocal sum is at least \(3\).
Example 7. Maximum on the boundary
Not every fixed-sum extremum occurs at equal variables.
Problem. Let \(a,b,c\ge0\), \(a+b+c=5\). Find the maximum of \(a^2+b^2+c^2\).
Since \(a^2\le5a\), \(b^2\le5b\), \(c^2\le5c\), we get \(\sum a^2\le5(a+b+c)=25\). Equality occurs at \(5,0,0\) and permutations.
Example 8. Two fixed quantities
If \(a+b+c\) and \(ab+bc+ca\) are known, the sum of squares is already determined.
Problem. If \(a+b+c=3\), \(ab+bc+ca=2\), find \(a^2+b^2+c^2\).
\(a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=9-4=5\).
Chapter
Hard Inequality Problems
Key idea
A hard inequality problem is rarely solved by one standard formula. Usually two or three moves must be combined: normalization, Cauchy, AM-GM, Jensen, UVW, substitution, or a local estimate that later telescopes.
Basic facts
The main tools in this module are Engel form, AM-GM with chosen factors, Jensen for convex and concave functions, Schur/UVW for symmetric expressions, and SOS ideas: reducing a difference to a sum of squares or a product of nonnegative factors.
When to use this method
If a standard method gives a weak estimate, look for a preliminary substitution or an additional local estimate. If the expression is cyclic with roots, try proving one inequality for a pair of variables and summing it. If several numbers have distance conditions, use the sum of squared pairwise differences.
How to recognise the method
Symmetry suggests UVW; fractions with squares suggest Cauchy; fixed sum and roots suggest Jensen; products and powers suggest AM-GM; distance conditions suggest pairwise differences; expressions like \((x-y)\sqrt{\cdots}\) suggest comparing the root with a linear expression.
Typical mistakes
A common mistake is applying a strong theorem without checking the domain or direction. Another is obtaining a true but too weak estimate. In hard problems, it is important to know in advance what should telescope or where the equality case should appear.
Mini-checklist
1. Is there a normalization? 2. Which standard method gives the first step? 3. Where is the equality case hidden? 4. Is a local estimate needed? 5. Does the result sum to the required form? 6. Is the obtained estimate strong enough?
Example 1. Cauchy with fixed denominator sum
Problem. If \(a+b+c=3\), prove \(\sum\frac{a^2}{1+a}\ge\frac32\).
By Cauchy, \(\sum\frac{a^2}{1+a}\ge\frac{(a+b+c)^2}{3+a+b+c}=\frac{9}{6}=\frac32\).
Example 2. Jensen after normalization
Problem. If \(a+b+c=3\), prove \(\sum\frac1{a^2+a+1}\ge1\).
The function \(f(x)=1/(x^2+x+1)\) is convex for \(x\ge0\). Jensen gives \(\frac13\sum f(a)\ge f(1)=1/3\).
Example 3. UVW move
Problem. Prove \((a^2+b^2+c^2)^2\ge3abc(a+b+c)\).
We have \(a^2+b^2+c^2\ge ab+bc+ca=q\). Also \(q^2\ge3abc(a+b+c)\). Hence \((\sum a^2)^2\ge q^2\ge3abc(a+b+c)\).
Example 4. Schur as a mixed method
Problem. Prove \(\sum a^4+abc(a+b+c)\ge\sum_{\mathrm{sym}}a^3b\).
This is Schur degree \(4\). By UVW, it is enough to check \(b=c\) and the boundary: for \(b=c=1\), \(a=t\), the difference is \(t^2(t-1)^2\ge0\); on the boundary we get \((a-b)^2(a^2+ab+b^2)\ge0\).
Example 5. Local estimate with a root
Problem. Prove that if \(u,v>0\), then \((v-u)\sqrt{u^2+3v^2}\ge v^2-u^2\).
If \(v\ge u\), then \(\sqrt{u^2+3v^2}\ge u+v\). If \(v\le u\), then \(\sqrt{u^2+3v^2}\le u+v\), and the factor \(v-u\) is negative, so the sign reverses. In both cases the estimate follows.
Example 6. Pairwise differences
Problem. For \(x_1,\ldots,x_n\), prove \(\sum_{i Expanding \(\sum_{i
Example 7. Power squeeze
Problem. Prove for \(n>2\): \((n-2)^{n+2}(n+2)^{n-2} Since \(n-2
Example 8. Discriminant move
Problem. If \(Ax^2+Bx+C>0\) for all \(x\), what can be said about the discriminant?
We must have \(A>0\) and \(B^2-4AC<0\). In hard problems one often builds another quadratic with the same discriminant.
Chapter
Mixed Inequality Sets
Key idea
In a mixed set, the statement no longer tells the student which method to use. The main task is to recognise the shape of the expression: fractions, powers, roots, symmetry, constraints, an extremal case, or a hidden identity.
Basic facts
This module uses AM-GM, Cauchy in Engel form, Holder, local estimates, Schur, identities for pairwise differences, and equality-case checking. Fractions with a manageable denominator sum often suggest Cauchy; cyclic cubes often suggest a local estimate; symmetric powers often suggest Schur or power means.
When to use this method
If denominators are sums, first try Cauchy. If the expression is homogeneous, check normalization. If \(a^3\), \(a^5\), and symmetry appear, look for Schur. If several numbers have distance conditions, sum the squares of all pairwise differences.
How to recognise the method
The phrase “for all real \(x\)” often leads to the discriminant. Roots of the form \(\sqrt{u^2+3v^2}\) can be compared with \(u+v\). Inequalities involving \(a+b+c\) and \(abc\) often need AM-GM or Schur. If equality is expected at \(a=b=c\), look for an estimate that becomes equality there.
Typical mistakes
Do not apply the strongest familiar method mechanically. A common mistake is getting a true but too weak estimate. Another is forgetting that multiplying by a negative factor reverses the inequality. In mixed problems, always check whether the equality case matches the statement.
Mini-checklist
1. Is the expression homogeneous? 2. Is there a natural equality case? 3. Can Cauchy be applied to the fractions? 4. Is a local estimate available for one term? 5. Is Schur or a power mean needed? 6. Is there a hidden identity for pairwise differences?
Example 1. Direct Cauchy
This example recalls that the method is sometimes visible from the denominator sum.
Problem. Prove \(\sum\frac{a}{b+c}\ge\frac32\) for \(a,b,c>0\).
By Cauchy, \(\sum\frac{a}{b+c}=\sum\frac{a^2}{a(b+c)}\ge\frac{(a+b+c)^2}{2(ab+bc+ca)}\ge\frac32\).
Example 2. Local estimate
Sometimes one term must be replaced by a linear expression.
Problem. Prove \(\frac{x^3}{x^2+xy+y^2}\ge\frac{2x-y}{3}\) for \(x,y>0\).
After multiplying by the positive denominator, we need \(3x^3\ge(2x-y)(x^2+xy+y^2)\). The difference is \((x-y)^2(x+y)\ge0\).
Example 3. Tangent estimate
This move is useful when the sum of variables is fixed.
Problem. Prove for \(x\ge0\): \(\frac1{1+x^2}\ge\frac{2-x}{2}\).
The difference equals \(\frac{x(x-1)^2}{2(1+x^2)}\ge0\).
Example 4. Holder for fractions
A cube in the numerator often suggests Holder.
Problem. Prove \(\sum\frac{a^3}{b^2+bc+c^2}\ge\frac{a+b+c}{3}\).
By Holder, \(\sum\frac{a^3}{b^2+bc+c^2}\ge\frac{(a+b+c)^3}{2(a^2+b^2+c^2)+ab+bc+ca}\). Since \(3(a+b+c)^2\ge2(a^2+b^2+c^2)+ab+bc+ca\), the result follows.
Example 5. Schur as symmetry recognition
When powers are symmetric, it is not always necessary to expand everything by hand.
Problem. Prove \(a^3+b^3+c^3+3abc\ge\sum_{\mathrm{sym}}a^2b\).
This is Schur of degree \(3\): \(\sum a(a-b)(a-c)\ge0\). Expanding gives exactly the stated inequality.
Example 6. Pairwise differences
A distance condition between numbers is rarely used one pair at a time.
Problem. For \(x_1,\ldots,x_n\), prove \(\sum_{i Expand the squares: each \(x_i^2\) appears \(n-1\) times, and the mixed terms give \(-2\sum_{i
Example 7. Root with sign direction
Here the sign of the multiplier must not be lost.
Problem. Prove \((v-u)\sqrt{u^2+3v^2}\ge v^2-u^2\) for \(u,v>0\).
If \(v\ge u\), then \(\sqrt{u^2+3v^2}\ge u+v\). If \(v\le u\), then \(\sqrt{u^2+3v^2}\le u+v\), but \(v-u\le0\), so the direction reverses after multiplication. In both cases \((v-u)\sqrt{u^2+3v^2}\ge(v-u)(u+v)\).
Example 8. Discriminant move
The phrase “positive for all \(x\)” almost always carries discriminant information.
Problem. If \(Ax^2+Bx+C>0\) for all real \(x\), prove \(Cx^2+Bx+A>0\) for all real \(x\).
We have \(A>0\), \(C=P(0)>0\), and the discriminant \(B^2-4AC<0\). The reversed quadratic has leading coefficient \(C>0\) and the same discriminant, so it is also positive for all \(x\).
Chapter
Mock Olympiads: Inequalities
Key idea
A mock set tests not one topic, but the ability to choose a method. One set may contain Cauchy, AM-GM, Holder, Schur, local estimates, discriminants, and identities for pairwise differences.
Basic facts
Before starting a set, recall five basic moves: a fraction with a manageable denominator sum often uses Cauchy; products and fixed products use AM-GM; cubes in numerators suggest Holder or a local estimate; symmetric powers suggest Schur; “for all \(x\)” conditions suggest the discriminant.
When to use this method
If a problem looks short, first look for a standard move. If the standard move gives a weak estimate, try a local replacement or normalization. If many variables have distance constraints, use the sum of squared pairwise differences.
How to recognise the method
Equality at \(a=b=c\) often points to a symmetric estimate. Equality at boundary values often requires a constraint or boundary analysis. If the expression is homogeneous, one quantity can often be normalized; if it is not homogeneous, first look for a fixed sum or product.
Typical mistakes
Do not start with the hardest problem in the set. Do not use an inequality without checking direction and domain. Do not leave a solution at “obvious by AM-GM”: an olympiad solution should show exactly which terms are being compared.
Mini-checklist
1. Find the easiest entry point. 2. Mark the likely equality case. 3. Check Cauchy for fractions. 4. Check AM-GM for products. 5. Recall Schur for symmetric powers. 6. After solving, check whether the estimate was strong enough.
Example 1. Quick entry point
Problem. Prove \(\sum\frac{a^2}{ab+ac+bc}\ge1\).
By Cauchy, \(\sum\frac{a^2}{ab+ac+bc}\ge\frac{(a+b+c)^2}{3(ab+bc+ca)}\ge1\).
Example 2. Local estimate
Problem. Prove \(\frac{x}{x^2+x+1}\le\frac13\).
The inequality is equivalent to \(3x\le x^2+x+1\), that is, \((x-1)^2\ge0\).
Example 3. Holder
Problem. Prove \(\sum\frac{a^3}{b^2+bc+c^2}\ge\frac{a+b+c}{3}\).
By Holder, the left side is at least \(\frac{(a+b+c)^3}{2\sum a^2+\sum ab}\), and the denominator is at most \(3(a+b+c)^2\).
Example 4. Schur
Problem. Prove \(\sum a^5+abc\sum a^2\ge\sum_{\mathrm{sym}}a^4b\).
This is Schur: \(\sum a^3(a-b)(a-c)\ge0\). Expanding gives the required form.
Example 5. Discriminant
Problem. If \(Ax^2+Bx+C>0\) for all \(x\), prove \(B^2<4AC\).
A quadratic with positive leading coefficient and a real root could not be positive everywhere. Hence the discriminant is negative.
Example 6. Pairwise differences
Problem. If \(y_1<\cdots Immediately, \(\sum_{i