Book 2. Olympiad Inequalities

We have \[2(a^2+b^2+c^2-ab-bc-ca)=(a-b)^2+(b-c)^2+(c-a)^2\ge0.\] Hence the inequality is true. Equality occurs exactly when \(a=b=c\).

\[x^2+y^2=\frac{(x+y)^2+(x-y)^2}{2}=98+\frac{(x-y)^2}{2}\ge98.\] The minimum is \(98\), attained at \(x=y=7\).

Since \(t^2+t+1>0\), we may multiply by \(3(t^2+t+1)\). We get \(3t\le t^2+t+1\), i.e. \((t-1)^2\ge0\). Equality occurs at \(t=1\).

Write the elements as \(a_1

We have \(up+vq-uq-vp=u(p-q)-v(p-q)=(u-v)(p-q)\). Both factors are positive, so the difference is positive.

Cyclically rename the variables so that \(x\ge z\). Then \(y+z\le y+x\), hence \(\frac{y}{y+z}\ge\frac{y}{y+x}\). Therefore \[\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}\ge\frac{x}{x+y}+\frac{y}{y+x}+\frac{z}{z+x}=1+\frac{z}{z+x}>1.\]

Adding gives \(-x-y>0\), so \(1-x-y>1>0\). Now multiply the inequalities \(x^2-x>y^2\) and \(y^2-y>x^2\). This is legitimate because \(x^2\) and \(y^2\) are positive, and the left sides are larger than them. We obtain \((x^2-x)(y^2-y)>x^2y^2\). Expanding and cancelling gives \(xy(1-x-y)>0\). Since \(1-x-y>0\), we get \(xy>0\).

Since \(a+b+c=0\), \[0=(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca).\] Therefore \[ab+bc+ca=-\frac{a^2+b^2+c^2}{2}\le0.\] Equality is possible only when \(a=b=c=0\).

Assume the contrary. Then for each \(i\), the discriminant \(4(p_i^2-q_i)\le0\), so \(p_i^2\le q_i\). Hence all \(q_i\ge0\), and the inequalities may be multiplied: \((p_1p_2p_3)^2\le q_1q_2q_3\). Thus \(N^2\le N\), impossible for \(N>1\). Therefore for some \(i\), \(p_i^2>q_i\), so the corresponding trinomial has two distinct real roots.

Let \(a_1\le a_2\le\cdots\le a_{2027}\). Since \(a_1+a_2+a_3\) is written on the board, it is at least \(a_1\), hence \(a_2+a_3\ge0\). Similarly, \(a_{2025}+a_{2026}+a_{2027}\) is written and is at most \(a_{2027}\), hence \(a_{2025}+a_{2026}\le0\). But \(a_2\le a_3\le a_{2025}\le a_{2026}\), so \[0\le a_2+a_3\le a_{2025}+a_{2026}\le0.\] Thus \(a_2+a_3=a_{2025}+a_{2026}=0\), and the ordering gives \(a_2=a_3=\cdots=a_{2026}=0\). Hence there are at least \(2025\) zeros. The example \(-1,0,\ldots,0,1\) shows the bound is sharp.

From \(a^3-b^3=3\) we get \(a>b\). Then \[3(a^2+b^2)=(a^2+b^2)(a^3-b^3)=a^5-b^5+a^2b^2(a-b).\] The right side is at least \(9\), because \(a^5-b^5\ge9\) and \(a^2b^2(a-b)\ge0\). Hence \(3(a^2+b^2)\ge9\), so \(a^2+b^2\ge3\).

Let \(a_1>a_2>\cdots>a_{12}>0\). Then \(a_1a_2a_3>a_4a_5a_6\). If \(a_6\ge1\), then \(a_1a_2a_3>a_4a_5a_6\ge a_4a_5\), so three numbers beat two. If \(a_6<1\), then \(a_7<1\), and \(a_1a_2a_3>a_4a_5a_6>a_4a_5a_6a_7\), so three numbers beat four.

Let the numbers be \(a_1,\ldots,a_{2n}\), with indices modulo \(2n\), and let \(S_i=a_i+a_{i+1}+\cdots+a_{i+n-1}\). The total sum is \(S>0\), so \(S_i+S_{i+n}=S\). Hence some \(S_i\) is positive. If all \(S_i>0\), any number works. Otherwise, while moving around the circle, there is a transition \(S_i>0\), \(S_{i+1}\le0\). Then the length-\(n\) arc ending at \(a_i\) from the other side has sum \(S-S_{i+1}>0\). Thus \(a_i\) is an endpoint of two arcs with positive sums.

Let \(m=37!\). Then \(38!=38m\), and \((38!)!=(38m)!\). Split \(1\cdot2\cdots38m\) into \(38\) blocks of length \(m\): in the \(j\)-th block, every factor is at most \(jm\), and at least one factor is smaller than \(jm\). Hence the product of the \(j\)-th block is strictly less than \((jm)^m\). Multiplying for \(j=1,\ldots,38\), \[(38m)!

If \(c\le950\), change \(c\) to \(0\). Then \(ax^2+bx=x(ax+b)\) has the integer root \(0\), and the cost is at most \(950\).

Now suppose \(c>950\). Then \(a+b<850\). Choose a square \(k^2\) nearest to \(c\). Since \(c<1800<43^2\), we may take \(k\le42\), and the distance from \(c\) to the nearest square is at most half the gap between neighbouring squares, hence at most \(42\). Change \(a\) to \(-1\), \(b\) to \(0\), and \(c\) to \(k^2\). The cost is at most \((a+1)+b+42<893\). The resulting trinomial \(-x^2+k^2\) has integer roots \(\pm k\). In both cases, \(950\) coins are enough.

Let \(M\) be the largest number. It cannot equal the absolute difference of its neighbours, because both neighbours are positive and at most \(M\). Hence \(M\) is the unique exception.

Choose a minimum \(d\). On one arc from \(d\) to \(M\), write \(d=a_0,a_1,\ldots,a_k=M\). By induction, \(a_{i+1}\ge a_i\), and since \(a_i=|a_{i+1}-a_{i-1}|\), in fact \(a_{i+1}=a_i+a_{i-1}\) for \(1\le i

The number \(d\) is not the exception, so \(d=|a_1-b_1|\). Suppose \(b_1>a_1\). Then \(b_1=a_1+d=a_2\). Continuing both recurrences gives \(a_0=b_0\le a_1

We need to prove \(x^9\ge16y^3\). From the condition, \(y^3\le x^5-4x\), so it is enough to show \(x^9\ge16x^5-64x\). But \[x^9-16x^5+64x=x(x^4-8)^2\ge0,\] since \(x>0\). Therefore \(x^9\ge16y^3\), hence \(x^3\ge\sqrt[3]{16}\,y\).

Assume \(A\cap B=\varnothing\). Then \(A\cup B\) contains \(2n\) distinct positive integers, so the sum of all its elements is at least \(1+2+\cdots+2n=n(2n+1)\). On the other hand, this sum equals \(n^2+n^2=2n^2\). But \(n(2n+1)>2n^2\), a contradiction. Hence a common element exists.

Adding gives \(x+y<0\). Suppose \(xy<0\). Without loss of generality, \(x>0\), \(y<0\). Then \(x+y<0\) implies \(00\). Hence \(xy\) is not negative; since \(x,y\ne0\), we get \(xy>0\).

Let the roots be \(r,s\), and suppose \(|r|\le1\), \(|s|>1\), since exactly one root lies in \([-1,1]\). By Vieta, \(rs=b>0\), so the roots have the same sign and \(b=|r||s|\). Since \(|s|>1\), we get \(|r|b\). Thus \(s\notin(-b,b)\). Therefore exactly one root lies in \((-b,b)\).

Call \(i\ge2\) bad if \(a_{i-1}a_{i+1}\) or \(a_{i-1}>a_i

Assume there are infinitely many bad indices, and choose one bad index \(k\). Take any \(n>k\). By the condition, \(a_n\) belongs to a monotone segment of length \(n+1\). Such a segment cannot contain all three terms \(a_{k-1},a_k,a_{k+1}\), because the direction changes there. Since the segment is so long and contains \(a_n\), it cannot start early enough to include \(a_{k-1}\); hence it must contain \(a_n,a_{n+1},a_{n+2}\). Thus \(n+1\) is not bad. Therefore for every \(n>k\), the index \(n+1\) is not bad, contradicting infinitely many bad indices. Hence there are only finitely many bad indices, and the tail is monotone.

For \(n=150,k=75\), we get \(a_{150}-a_{75}\ge150^3-75^3\), so \(a_{150}\ge150^3-75^3\). For \(n=75,k=150\), we get \(a_{75}-a_{150}\ge75^3-150^3\), so \(a_{150}\le150^3-75^3\). Therefore \[a_{150}=150^3-75^3=7\cdot75^3.\]

For every \(t\), \(t^2+t+1=\left(t+\frac12\right)^2+\frac34\ge\frac34\). Hence \(\frac1{t^2+t+1}\le\frac43\), with equality only at \(t=-\frac12\). The sum of three such fractions is at most \(4\). If it equals \(4\), each fraction must equal \(\frac43\), so \(x=y=z=-\frac12\), impossible for distinct numbers. Answer: no.

The number \(p_1=x_1-\frac1{x_1}\) is not a positive integer: if \(x_1=1\), it is \(0\), and if \(x_1>1\), it is not an integer. Hence at most \(11\) of the \(p_i\) can be positive integers.

We show that \(11\) is attainable. Take \(x_i=i+1\). Then \[x_i-\frac1{x_i}=i+1-\frac1{i+1}=\frac{i(i+2)}{i+1}.\] Therefore \[p_i=\frac{1\cdot3}{2}\cdot\frac{2\cdot4}{3}\cdots\frac{i(i+2)}{i+1}=\frac{i!(i+2)}2.\] For \(i\ge2\), this is a positive integer. Thus \(11\) of \(p_1,\ldots,p_{12}\) can be positive integers, and more is impossible.

From \(a^2+b^2<(a-b)^2=a^2-2ab+b^2\), we get \(ab<0\). Similarly, \(bc<0\). Thus \(a\) and \(b\) have opposite signs, and \(b\) and \(c\) have opposite signs; hence \(a\) and \(c\) have the same sign, so \(ac>0\). Then \[(a+c)^4-a^4-c^4=4a^3c+6a^2c^2+4ac^3=2ac(2a^2+3ac+2c^2)>0,\] because \(ac>0\) and \(2a^2+3ac+2c^2>0\). Therefore \(a^4+c^4<(a+c)^4\).

Answer: \(k(k-1)\). Construction: take one isolated vertex, split the remaining vertices into sets \(X\) and \(Y\) of sizes \(k\) and \(k-1\), and connect every vertex of \(X\) to every vertex of \(Y\). Then vertices in \(X\) have degree \(k-1\), vertices in \(Y\) have degree \(k\), the condition holds, and there are \(k(k-1)\) edges.

Now prove the upper bound. Let \(X_i\) be the set of vertices of degree \(i\), and let \(m\) be the maximum degree. If \(m\le k-1\), then \(2E\le2k(k-1)\), so \(E\le k(k-1)\).

If \(m\ge k+1\), choose a vertex of degree \(m\). All its neighbours have degree \(m-1\), so \(|X_{m-1}|\ge m\). Any vertex in \(X_{m-1}\) has \(m-1\) neighbours in \(X_m\cup X_{m-2}\), so the classes \(X_{m-1},X_m,X_{m-2}\) contain at least \(m+(m-1)>2k\) vertices, impossible.

It remains to consider \(m=k\). Edges connect only adjacent degree classes, so the graph is bipartite with parts \(Y=X_k\cup X_{k-2}\cup\cdots\) and \(Z=X_{k-1}\cup X_{k-3}\cup\cdots\). On one hand, \(E\le k|Y|\); on the other, \(E\le(k-1)|Z|\). If \(|Y|\le k-1\), then \(E\le k(k-1)\). If \(|Y|\ge k\), then \(|Z|\le k\), so \(E\le(k-1)k\). The upper bound is proved.

If the number of negative entries were even, there would be a positive number \(a\), and the product of all the other numbers would be positive. Then \(a+\prod_{\ne a}>0\), a contradiction. Hence the number of negative entries is odd.

Partition the numbers into two groups. Exactly one of the two products is negative. Let \(X=x_1x_2\cdots x_r<0\), \(Y=y_1y_2\cdots y_s>0\), and choose \(x_1<0\). Then \(x_2\cdots x_r>0\), and since the numbers are integers, \(x_2\cdots x_r\ge1\). Therefore \[X+Y=x_1(x_2\cdots x_r)+Y\le x_1+Y\le x_1+Y(x_2\cdots x_r).\] The last expression is negative by the condition applied to \(x_1\). Hence \(X+Y<0\).

Put \(t=\sqrt[2^{n+1}]{a}\). Then \(t\ne1\), \(\sqrt[2^n]{a}=t^2\), and \[x_n=2^n(t^2-1),\qquad x_{n+1}=2^{n+1}(t-1).\] Hence \[x_n-x_{n+1}=2^n(t^2-1)-2^{n+1}(t-1)=2^n(t-1)^2>0.\] Thus \(x_n>x_{n+1}\) for every \(n\), so the sequence is strictly decreasing.

For the pair \((i,j)\), the condition gives \(u_i-u_j\ge i^2-j^2\). For the pair \((j,i)\), it gives \(u_j-u_i\ge j^2-i^2\), i.e. \(u_i-u_j\le i^2-j^2\). The lower and upper bounds coincide, so \(u_i-u_j=i^2-j^2\).

Call a card uniform if both of its sides are greater than \(1\) or both are less than \(1\). Initially there is a uniform card. If a uniform card is replaced by its reciprocal, it remains uniform: large-large becomes small-small and conversely.

Now suppose a uniform card \((a,b)\) is split by the second operation. If \(a>1,b>1\), then from \(c\cdot\frac ac=a>1\), at least one of \(c,\frac ac\) is greater than \(1\); together with \(b>1\), the corresponding new card is uniform. If \(a<1,b<1\), then at least one of \(c,\frac ac\) is less than \(1\), otherwise their product would be at least \(1\); again a uniform card appears. Thus uniform cards cannot disappear completely. The desired final collection contains none, a contradiction.

By AM-GM, \(x+\frac{36}{x}\ge2\sqrt{x\cdot\frac{36}{x}}=12\). Equality occurs at \(x=6\).

\(\frac{a+b+c}{3}\ge\sqrt[3]{abc}=3\), hence \(a+b+c\ge9\). Equality occurs when \(a=b=c=3\).

By AM-GM, \(\sqrt[3]{abc}\le\frac{a+b+c}{3}=5\). Cubing gives \(abc\le125\).

\(\frac{x+x+y}{3}\ge\sqrt[3]{x^2y}=4\). Therefore \(2x+y\ge12\). Equality occurs at \(x=y=4\).

Using the standard AM-GM/Cauchy consequence \((a+b+c)\left(\frac1a+\frac1b+\frac1c\right)\ge(1+1+1)^2=9\), and \(a+b+c=1\), we get the result.

The sum of \(1+a,1+b,1+c\) is \(9\). By AM-GM, their product is at most \(\left(\frac93\right)^3=27\). Equality occurs at \(a=b=c=2\).

By AM-GM for \(x,x,y,z\): \(\frac{x+x+y+z}{4}\ge\sqrt[4]{x^2yz}=2\). Hence \(2x+y+z\ge8\).

By AM-GM, \[\frac{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}{3}\ge\sqrt[3]{\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{c}{a}}=1.\] The result follows.

By Cauchy, \[\left(\sum\sqrt{\frac{a}{b+c}}\right)^2\ge\frac{(a+b+c)^2}{a(b+c)+b(c+a)+c(a+b)}.\] Since the denominator is \(2(ab+bc+ca)\le\frac{2}{3}(a+b+c)^2=\frac23\), the sum is at least \(\sqrt{\frac32}\). This standard bound illustrates the same preparation principle.

By AM-GM, \(1+a\ge2\sqrt a\), \(1+b\ge2\sqrt b\), \(1+c\ge2\sqrt c\). Multiplying gives \((1+a)(1+b)(1+c)\ge8\sqrt{abc}=8\). Equality occurs at \(a=b=c=1\).

By AM-GM, \(\frac{x+x+x+y+z}{5}\ge\sqrt[5]{x^3yz}=\sqrt[5]{32}=2\). Hence \(3x+y+z\ge10\).

Let \(P=abcd\), \(S=a+b+c+d\). If \(P\ge16\), then \[a^2+b^2+c^2+d^2\ge4\left(\frac{S}{4}\right)^2\ge4\left(\frac{P}{8}\right)^2=\frac{P^2}{16}\ge P.\] If \(P<16\), then by AM-GM \[a^2+b^2+c^2+d^2\ge4\sqrt[4]{a^2b^2c^2d^2}=4\sqrt P.\] Since \(P<16\), \(4\sqrt P>P\). The result follows in both cases.

We have \(cd\le |cd|\le\frac{c^2+d^2}{2}=\frac{9-a^2-b^2}{2}\). It is enough to prove \(2(3+a)(3+b)\ge9-a^2-b^2\). After moving all terms to the left, we get \(a^2+2ab+b^2+6a+6b+9=(a+b+3)^2\ge0\). Hence the inequality is true.

Since \(ab+bc+ca=2\), \[a+\frac2a=a+\frac{ab+ac+bc}{a}=a+b+c+\frac{bc}{a}.\] By AM-GM, \(a+\frac{bc}{a}\ge2\sqrt{bc}\). Hence \(a+\frac2a\ge b+c+2\sqrt{bc}=(\sqrt b+\sqrt c)^2\), so \(\sqrt{a+\frac2a}\ge\sqrt b+\sqrt c\). Adding the three cyclic estimates gives the result.

Since \(|u_i|>1\), the numbers \(x_i=\frac{u_i+1}{u_i-1}\) are positive. Their product is \(1\). Also, \(\frac1{u_i-1}=\frac{x_i-1}{2}\). Thus it is enough to prove \(x_1+x_2+x_3+x_4>4\). By AM-GM, \(x_1+x_2+x_3+x_4\ge4\sqrt[4]{x_1x_2x_3x_4}=4\). Equality would require all \(x_i=1\), impossible for finite \(u_i\). Hence the sum is strictly greater than \(4\), and the required sum is positive.

After multiplying, it is enough to prove \[a^2b^2c^2+a^2b^2d^2+a^2c^2d^2+b^2c^2d^2\le1.\] Assume \(a\ge b\ge c\ge d\). By AM-GM for \(a,b,c+d\), \(ab(c+d)\le\left(\frac{a+b+c+d}{3}\right)^3=1\), so \(a^2b^2(c+d)^2\le1\). It remains to check that the left side is at most \(a^2b^2(c+d)^2\). Expanding, this reduces to \(c^2d^2(a^2+b^2)\le2a^2b^2cd\), which follows from \(cd\le b^2\) and \(cd\le a^2\). The original inequality follows.

We may assume \(x\ge y\ge z\). Let \(u=x-y\ge0\), \(v=y-z\ge0\), so \(x-z=u+v\), and the left side is \(uv(u+v)\). From \(x^2+y^2+z^2=1\), we get \((x-y)^2+(y-z)^2+(x-z)^2\le3\), hence \(u^2+v^2+(u+v)^2\le3\). For fixed \(s=u+v\), \(uv\le\frac{s^2}{4}\). Also \(u^2+v^2\ge\frac{s^2}{2}\), so \(\frac32s^2\le3\), i.e. \(s\le\sqrt2\). Therefore \(uv(u+v)\le\frac{s^3}{4}\le\frac{(\sqrt2)^3}{4}=\frac1{\sqrt2}\).

For each \(i\), \[(1+x_i^2)(1+x_{i+1}^2)-(1+x_ix_{i+1})^2=(x_i-x_{i+1})^2\ge0.\] Multiplying over all \(i\), we obtain \[\prod_{i=1}^{n}\frac{1+x_i^2}{1+x_ix_{i+1}}\ge1.\] Applying AM-GM to the \(n\) positive fractions, their sum is at least \(n\) times the \(n\)-th root of the product, hence at least \(n\).

We have \(b+c\ge2\sqrt{bc}\), hence \[4\frac{\sqrt{ab-1}}{b+c}\le2\sqrt{\frac{ab-1}{bc}}=2\sqrt{\left(a-\frac1b\right)\frac1c}\le a-\frac1b+\frac1c.\] Similarly, \[4\frac{\sqrt{bc-1}}{c+a}\le b-\frac1c+\frac1a,\qquad 4\frac{\sqrt{ca-1}}{a+b}\le c-\frac1a+\frac1b.\] Adding gives that four times the right side is at most \(a+b+c\), which is the desired inequality.

Answer: \(\frac1{76}\). Peter takes \(\frac12,\frac1{38},\ldots,\frac1{38}\). The number \(\frac12\) must be paired with \(\frac1{38}\), so the largest product is at least \(\frac1{76}\), and all other products are no larger.

Now let the numbers be arbitrary, \(x_1\ge\cdots\ge x_{20}\). Basil pairs \(x_k\) with \(x_{21-k}\). For \(k\le10\), put \(P=x_kx_{21-k}\). From \(kx_k\le x_1+\cdots+x_k=S\) and \((21-2k)x_{21-k}\le1-S\), \[P\le\frac{S(1-S)}{k(21-2k)}\le\frac1{4k(21-2k)}\le\frac1{76}.\] Thus Basil can guarantee that the maximum is at most \(\frac1{76}\). The value is proved.

Lower bound: take \(0,\frac12,\frac1{4(n-1)},\ldots,\frac1{4(n-1)}\). The number \(\frac12\) has two neighbours; at least one of them is \(\frac1{4(n-1)}\), so the maximum product is at least \(\frac1{8(n-1)}\).

For the upper bound, order \(x_1\ge\cdots\ge x_{2n}\). Arrange the numbers so that all potentially large neighbouring products have the form \(x_kx_{2n-k}\), \(1\le k\le n-1\). As in the pairing problem, if \(S=x_1+\cdots+x_k\), then \[x_kx_{2n-k}\le\frac{S(1-S)}{k(2n-2k)}\le\frac1{4k(2n-2k)}\le\frac1{8(n-1)}.\] Hence the bound is attainable for all choices and is sharp.

Since the sequence is nonincreasing, \[a_1^2+\cdots+a_{16}^2\ge4a_7^2+4a_{10}^2+4a_{13}^2+4a_{16}^2.\] Hence \[1\ge4(a_7^2+a_{10}^2+a_{13}^2+a_{16}^2)\ge(a_7+a_{10}+a_{13}+a_{16})^2,\] by RMS-AM in the last step. Thus the first sum is at most \(1\). Similarly, \(1\ge3(a_7^2+a_8^2+a_9^2)\ge(a_7+a_8+a_9)^2\), which gives the second inequality.

Subtracting gives \((a-b)(b-c)(c-a)=2\). Let \(c\) be the largest number. Then the signs imply \(a0\), \(y=c-b>0\), \(z=c-a=x+y\). Then \(xyz=2\). By AM-GM, \(4xy\le(x+y)^2=z^2\), hence \(2=xyz\le\frac{z^3}{4}\), so \(z\ge2\). If \(z=2\), equality in \(4xy\le z^2\) gives \(x=y=1\). Then \(b=a+1>1\), \(c=a+2>2\), and \(a^2b+b^2c+c^2a>b^2c>2\), contradicting the first equality. Therefore \(z>2\), i.e. \(c-a>2\).

Add the first two inequalities: \(2x+2y>x^2+y^2+2z^2\). Rearranging, \[0>(x-1)^2+(y-1)^2+2(z^2-1).\] Since the first two terms are nonnegative, \(z^2<1\), hence \(z<1\). Similarly, adding the other pairs gives \(x<1\) and \(y<1\).

Multiplying the three factors by \(\frac ba,\frac cb,\frac ac\), we get \[\left(b-\frac1a\right)\left(c-\frac1b\right)\left(a-\frac1c\right)=1.\] All three factors are positive. By AM-GM, \[\left(b-\frac1a\right)^2+\left(c-\frac1b\right)^2+\left(a-\frac1c\right)^2\ge3.\] Expanding gives \[a^2+b^2+c^2+\frac1{a^2}+\frac1{b^2}+\frac1{c^2}\ge3+2\left(\frac ba+\frac cb+\frac ac\right).\] Also, \(\frac ba+\frac cb+\frac ac\ge3\). Hence the right side is at least \(6+\frac ba+\frac cb+\frac ac\). Expanding the left side of the desired inequality gives \(a^2+b^2+c^2+\frac1{a^2}+\frac1{b^2}+\frac1{c^2}-6\). The result follows.

By Cauchy for \((a,b)\) and \((c,d)\), \((a^2+b^2)(c^2+d^2)\ge(ac+bd)^2\). After expansion, the difference is \((ad-bc)^2\ge0\).

By Engel form, \(\frac{x^2}{p}+\frac{y^2}{q}\ge\frac{(x+y)^2}{p+q}\). Equality occurs when \(\frac{x}{p}=\frac{y}{q}\).

By Cauchy, \[9=(1+1+1)^2\le(a+b+c)\left(\frac1a+\frac1b+\frac1c\right)=12\left(\frac1a+\frac1b+\frac1c\right).\] The result follows.

\[\sum\frac{a}{b+c}=\sum\frac{a^2}{a(b+c)}\ge\frac{(a+b+c)^2}{a(b+c)+b(c+a)+c(a+b)}=\frac{(a+b+c)^2}{2(ab+bc+ca)}\ge\frac32,\] since \((a+b+c)^2\ge3(ab+bc+ca)\).

By Cauchy, \[\sum\frac{x^2}{x+y}\ge\frac{(x+y+z)^2}{(x+y)+(y+z)+(z+x)}=\frac{x+y+z}{2}.\]

By Cauchy, the left side is at least \[\frac{(x+y+z)^2}{2(x^2+y^2+z^2)+xy+yz+zx}.\] Since \(2(x^2+y^2+z^2)+xy+yz+zx\le2(x+y+z)^2\), the left side is at least \(\frac12\).

By Cauchy, \[\sum_{i=1}^{n}\frac{1^2}{1+x_i}\ge\frac{(1+\cdots+1)^2}{(1+x_1)+\cdots+(1+x_n)}=\frac{n^2}{n+\sum x_i}.\]

By Cauchy, \[\frac{4}{a}+\frac{9}{b}+\frac{16}{c}\ge\frac{(2+3+4)^2}{a+b+c}=\frac{81}{6}=\frac{27}{2}.\] Equality occurs when \(\frac{2}{a}=\frac{3}{b}=\frac{4}{c}\), i.e. \(a:b:c=2:3:4\). This is attainable with sum \(6\), so the minimum is \(\frac{27}{2}\).

By Cauchy, \[\sum\frac{a^2}{a+2b}\ge\frac{(a+b+c)^2}{(a+2b)+(b+2c)+(c+2a)}=\frac{(a+b+c)^2}{3(a+b+c)}=\frac{a+b+c}{3}.\]

By Cauchy, the left side is at least \[\frac{(a+b+c)^2}{2(a^2+b^2+c^2)+ab+bc+ca}.\] Since \(2(a^2+b^2+c^2)+ab+bc+ca\le2(a+b+c)^2\), we get the lower bound \(\frac12\).

By Cauchy, \[\sum\frac{a^2}{b+2c}\ge\frac{(a+b+c)^2}{(b+2c)+(c+2a)+(a+2b)}=\frac{(a+b+c)^2}{3(a+b+c)}=\frac{a+b+c}{3}.\]

By Cauchy, \[\frac1x+\frac4y+\frac9z\ge\frac{(1+2+3)^2}{x+y+z}=\frac{36}{10}=\frac{18}{5}.\] Equality occurs when \(1/x=2/y=3/z\), i.e. \(x:y:z=1:2:3\). This is attainable with sum \(10\), so the minimum is \(\frac{18}{5}\).

By Cauchy, the left side is at least \[\frac{(a+b+c)^2}{2(a^2+b^2+c^2)+ab+bc+ca}.\] This denominator is at most \(2(a+b+c)^2\), so the left side is at least \(\frac12\). The condition \(abc=1\) shows that the problem may look multiplicative, but the method is homogeneous.

By Cauchy, the left side is at least \[\frac{(a+b+c)^2}{(a+b+c)+(a+2b+c)+(a+b+3c)}=\frac{(a+b+c)^2}{3a+4b+5c}.\]

By Cauchy, \[(\sqrt{u+v}+\sqrt{v+w}+\sqrt{w+u})^2\le3((u+v)+(v+w)+(w+u))=6(u+v+w)=36.\] Hence the sum of roots is at most \(6\).

For each term, \[\frac{a^3}{a^2+b+c}=a-\frac{a(b+c)}{a^2+b+c}\ge a-\frac{a(b+c)}{2a\sqrt{b+c}}=a-\frac{\sqrt{b+c}}2.\] Adding the four estimates and using \(a+b+c+d=10\) gives the result.

As in the previous problem, \(\frac{a^3}{a^2+b+c}\ge a-\frac{\sqrt{b+c}}2\), and similarly for the other terms. Thus the left side is at least \[8-\frac12(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+d}+\sqrt{d+a}).\] It remains to show that the root sum is at most \(8\). By Cauchy, \[(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+d}+\sqrt{d+a})^2\le4\cdot2(a+b+c+d)=64.\] Hence the root sum is at most \(8\), and the left side is at least \(4\).

\[\frac{a}{b^4+2b}=\frac{a}{b}-\frac{2ab^2}{b^3+2}\ge\frac{a}{b}-\frac{2ab^2}{3b}=\frac{a}{b}-\frac23ab,\] because \(b^3+2=b^3+1+1\ge3b\). Adding cyclically, \[\sum\frac{a}{b^4+2b}\ge\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)-\frac23(ab+bc+ca).\] The first bracket is at least \(3\) by AM-GM. Also, \(a+b+c=3\) implies \(ab+bc+ca\le3\). Thus the right side is at least \(3-\frac23\cdot3=1\).

By Cauchy, the left side is at least \[\frac{(x^2+y^2+z^2)^2}{2(x^2+y^2+z^2)+xy+yz+zx}.\] Since \(xy+yz+zx\le x^2+y^2+z^2\), the denominator is at most \(3(x^2+y^2+z^2)\). This gives the required bound \(\frac{x^2+y^2+z^2}{3}\).

By Cauchy, \[\sum\frac{a^2}{2a+b+c}\ge\frac{(a+b+c)^2}{(2a+b+c)+(2b+c+a)+(2c+a+b)}=\frac{(a+b+c)^2}{4(a+b+c)}=\frac{a+b+c}{4}.\]

We have \(ax+by-ay-bx=(b-a)(y-x)\). Both factors are nonnegative, so the difference is nonnegative.

The difference between the right and left sides is \(ax+cz-az-cx=(c-a)(z-x)\ge0\). Hence the inequality holds.

By the one-swap inequality, \(bz+cx\le bx+cz\), since \(b\le c\) and \(x\le z\). Also \(ay+bx\le ax+by\), since \(a\le b\) and \(x\le y\). Adding gives the result.

Expanding the difference gives \[3\sum ax-\sum a\sum x=(b-a)(y-x)+(c-a)(z-x)+(c-b)(z-y).\] Every term on the right is nonnegative because both sequences are increasing.

Order \(a,b,c\) as \(u_1\le u_2\le u_3\). By rearrangement, \(u_1^2+u_2^2+u_3^2\) is the largest among sums \(\sum u_i u_{\sigma(i)}\). The sum \(ab+bc+ca\) is one of these permutations, so it is not larger than the sum of squares.

If for \(i

After renaming, assume \(x\le y\le z\). Then \(x^2\le y^2\le z^2\). By Chebyshev, \[\frac{x^3+y^3+z^3}{3}\ge\frac{x+y+z}{3}\cdot\frac{x^2+y^2+z^2}{3}.\] Multiplying by \(9\) gives the result.

Order the numbers increasingly. Their squares have the same order. By Chebyshev, \[\frac{\sum a^3}{4}\ge\frac{\sum a}{4}\cdot\frac{\sum a^2}{4}.\] Multiplying by \(16\) gives the inequality.

Let \(u_1\le u_2\le u_3\) be \(a,b,c\) in increasing order. Then \(u_1^2\le u_2^2\le u_3^2\). By rearrangement, the largest sum \(\sum u_i^2u_{\sigma(i)}\) is \(\sum u_i^3\). The right side of the original inequality is one such permutation, so it is not larger.

Order \(a,b,c\) as \(u_1\le u_2\le u_3\). Then \(u_1^3\le u_2^3\le u_3^3\). By rearrangement, the largest product sum \(u_i^3\cdot u_{\sigma(i)}\) is \(u_1^4+u_2^4+u_3^4\). The right side is one of these permutations.

From the cube inequality, \[3\sum a^3\ge\left(\sum a\right)\left(\sum a^2\right).\] Also, \(\sum a^2\ge\frac{(\sum a)^2}{3}\). Therefore \[3\sum a^3\ge \sum a\cdot\frac{(\sum a)^2}{3}=\frac{(\sum a)^3}{3},\] so \(\sum a^3\ge\frac{(\sum a)^3}{9}\).

By Chebyshev, \[\frac{ax+by+cz}{3}\ge\frac{a+b+c}{3}\cdot\frac{x+y+z}{3}=0.\] Hence \(ax+by+cz\ge0\).

The sequences \(a,b,c\) and \(\frac1c,\frac1b,\frac1a\) are both increasing. Therefore \(\frac ac+\frac bb+\frac ca\) is the maximum among all permutations. The sum \(\frac ab+\frac ba+\frac cc\) is one such permutation, so it is not larger.

By rearrangement, the maximum is \(ax+by+cz\), and the minimum is \(az+by+cx\). If some elements among \(a,b,c\) or \(x,y,z\) are equal, several permutations may give the maximum or minimum.

Expanding the right side, the coefficient of \(x\) is \(-b+a-c+a=2a-b-c\), the coefficient of \(y\) is \(b-a-c+b=-a+2b-c\), and the coefficient of \(z\) is \(c-a+c-b=-a-b+2c\). This matches the left side. Since every difference on the right is nonnegative, the left side is nonnegative, which gives Chebyshev.

After ordering, the sequences \(u_i^3\) and \(u_i\) have the same order. By rearrangement, the largest product sum \(u_i^3\cdot u_{\sigma(i)}\) is \(\sum u_i^4\). The cyclic right side is one of the permutations, so it is not larger than \(\sum u_i^4\).

Since the sequences are similarly ordered, Chebyshev gives \[\frac1n\sum_{i=1}^n a_i b_i\ge\left(\frac1n\sum_{i=1}^n a_i\right)\left(\frac1n\sum_{i=1}^n b_i\right)=0.\] Hence \(\sum a_i b_i\ge0\).

After ordering, \(a^2,b^2,c^2\) and \(a^3,b^3,c^3\) increase together. By Chebyshev, \[\frac{a^5+b^5+c^5}{3}\ge\frac{a^2+b^2+c^2}{3}\cdot\frac{a^3+b^3+c^3}{3}.\] Multiplying by \(9\) gives the result.

Reorder the numbers so that \(x_1\le\cdots\le x_n\). Then \(x_1^m\le\cdots\le x_n^m\). Chebyshev gives \[\frac1n\sum x_i^{m+1}\ge\left(\frac1n\sum x_i^m\right)\left(\frac1n\sum x_i\right).\] Multiplying by \(n\) gives the result.

Order \(a,b,c\) as \(u_1\le u_2\le u_3\). Then \(u_i^4\) are ordered in the same way. The maximum of \(\sum u_i^4u_{\sigma(i)}\) is \(\sum u_i^5\). The right side is one of the permutations, so it is not larger.

After ordering \(a,b,c\), we have \(u_1^2\le u_2^2\le u_3^2\) and \(u_1^4\le u_2^4\le u_3^4\). By rearrangement, the largest sum \(u_i^4u_{\sigma(i)}^2\) is \(\sum u_i^6\). The right side corresponds to one permutation, so it is not larger than the left side.

By Chebyshev for \(a^2\) and \(a^3\), \[3\sum a^5\ge(\sum a^2)(\sum a^3).\] Also \(3\sum a^3\ge(\sum a)(\sum a^2)\), and \(\sum a^2\ge\frac{(\sum a)^2}{3}\). Hence \(\sum a^3\ge\frac{(\sum a)^3}{9}\). Substituting this and using \(\sum a^2\ge\frac{(\sum a)^2}{3}\) again, we get \[3\sum a^5\ge\frac{(\sum a)^2}{3}\cdot\frac{(\sum a)^3}{9}=\frac{(\sum a)^5}{27}.\] Therefore \(\sum a^5\ge\frac{(\sum a)^5}{81}\).

By Chebyshev, \[\frac1n\sum_{i=1}^n i\,x_i\ge\left(\frac1n\sum_{i=1}^n i\right)\left(\frac1n\sum_{i=1}^n x_i\right)=0.\] Therefore \(\sum i x_i\ge0\).

Reorder the numbers as \(u_1\le\cdots\le u_n\). Then \(u_1^m\le\cdots\le u_n^m\). By rearrangement, the maximum of \(\sum u_i^m u_{\tau(i)}\) is attained at \(\tau(i)=i\), and equals \(\sum u_i^{m+1}\). The right side of the original inequality corresponds to some permutation \(\tau\), so it is not larger.

The function \(x^2\) is convex. Jensen gives \(\frac{a^2+b^2+c^2}{3}\ge\left(\frac{a+b+c}{3}\right)^2\). Multiplying by \(3\) gives the result.

By Jensen for the concave function \(\sqrt{x}\), \[\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{3}\le\sqrt{\frac{x+y+z}{3}}=\sqrt{9}=3.\] Hence the sum of roots is at most \(9\).

By Jensen, \[\frac13\left(\frac1a+\frac1b+\frac1c\right)\ge\frac{1}{(a+b+c)/3}=\frac{3}{a+b+c}.\] Multiply by \(3\).

Since \(\ln x\) is concave, \[\frac{\ln a+\ln b+\ln c}{3}\le\ln\frac{a+b+c}{3}=\ln2.\] Hence \(\ln(abc)\le3\ln2=\ln8\), so \(abc\le8\).

We have \(x^2-2x+1=(x-1)^2\ge0\). This says exactly that the graph of the convex function \(x^2\) lies above its tangent at \(1\).

Since \(x>0\), the inequality is equivalent to \(1\ge2x-x^2\), or \((x-1)^2\ge0\). This is the tangent to \(1/x\) at \(1\).

Since \(f(t)=t^2\) is convex, \[p f(x)+(1-p)f(y)\ge f(px+(1-p)y).\] This is exactly the required inequality.

By the concavity of \(\sqrt{x}\), \[\sqrt{\frac{x+2y}{3}}\ge\frac13\sqrt{x}+\frac23\sqrt{y}.\] Multiply by \(3\).

By Jensen, \[\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{3}\le\sqrt{\frac{48}{3}}=4.\] Hence the sum is at most \(12\). Equality occurs at \(x=y=z=16\), so the maximum is \(12\).

Since \(\ln x\) is concave, \[\frac{\ln a+\ln b+\ln c}{3}\le\ln\frac{a+b+c}{3}=\ln4.\] Therefore \(\ln(abc)\le\ln64\), so \(abc\le64\). Equality holds at \(a=b=c=4\).

By Jensen, \[\frac13\sum\frac{1}{1+a}\ge\frac{1}{1+(a+b+c)/3}=\frac{1}{3}.\] Multiplying by \(3\), we get the result.

The function \(f(x)=\frac{x}{1+x}\) is concave for \(x\ge0\). Jensen gives \[\frac13\sum\frac{a}{1+a}\le \frac{1}{1+1}=\frac12.\] Hence the sum is at most \(\frac32\).

By Jensen, \[\frac{a^4+b^4+c^4}{3}\ge\left(\frac{a+b+c}{3}\right)^4.\] Multiplying by \(3\) gives \(\sum a^4\ge\frac{(a+b+c)^4}{27}\).

By Jensen, \[\frac{e^x+e^y+e^z}{3}\ge e^{(x+y+z)/3}=e^0=1.\] Therefore the sum is at least \(3\).

The function \(\ln x\) is concave, so its graph lies below every tangent. At \(1\), the tangent has equation \(y=x-1\). Hence \(\ln x\le x-1\).

Since \(\sqrt{x}\) is concave, its graph lies below the tangent at \(1\). This tangent is \(y=1+\frac12(x-1)=\frac{x+1}{2}\). Hence \(\sqrt{x}\le\frac{x+1}{2}\).

The function \(\ln(1+t)\) is concave for \(t\ge0\). Thus \[\frac{\ln(1+x)+\ln(1+y)+\ln(1+z)}{3}\le\ln\left(1+\frac{x+y+z}{3}\right)=\ln2.\] Hence the logarithm of the product is at most \(\ln8\), so the product is at most \(8\).

The function \(f(t)=\frac1{2+t}\) is convex for \(t\ge0\). Jensen gives \[\frac13\sum\frac1{2+x}\ge \frac1{2+(x+y+z)/3}=\frac13.\] Multiply by \(3\).

By Jensen, \[\frac{x^4+y^4+z^4}{3}\ge\left(\frac{x+y+z}{3}\right)^4=1.\] Thus the sum of fourth powers is at least \(3\). Equality holds at \(x=y=z=1\).

The function \(f(t)=t\ln t\) is convex for \(t>0\). Jensen gives \[\frac{x\ln x+y\ln y+z\ln z}{3}\ge f\left(\frac{x+y+z}{3}\right)=f(1)=0.\] Therefore \(\ln(x^x y^y z^z)\ge0\), so \(x^x y^y z^z\ge1\).

The function \(e^x\) is convex on the whole real line. Therefore Jensen says that the average of the function values is at least the function value at the average argument: \[\frac1n\sum e^{x_i}\ge e^{(\sum x_i)/n}.\]

The function \(\ln(1+x)\) is concave for \(x\ge0\). Hence \[\frac1n\sum_{i=1}^n\ln(1+x_i)\le\ln\left(1+\frac{S}{n}\right).\] Multiplying by \(n\), we get \(\ln\prod(1+x_i)\le n\ln(1+S/n)\). Exponentiating gives the result.

Since \(f(t)=t\ln t\) is convex for \(t>0\), Jensen gives \[\frac1n\sum_{i=1}^n x_i\ln x_i\ge f\left(\frac{x_1+\cdots+x_n}{n}\right)=f(1)=0.\] Hence \(\ln\prod x_i^{x_i}=\sum x_i\ln x_i\ge0\). Therefore \(\prod x_i^{x_i}\ge1\).

By Jensen, \[\frac1n\sum_{i=1}^n\frac1{x_i^2}\ge \frac{1}{\left((x_1+\cdots+x_n)/n\right)^2}=1.\] Multiplying by \(n\), we get the result.

\(p^2=(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)=\sum a^2+2q\). Hence \(\sum a^2=p^2-2q\).

By the identity, \(\sum a^3-3r=p((p^2-2q)-q)=p(p^2-3q)=p^3-3pq\). Therefore \(\sum a^3=p^3-3pq+3r\).

Expanding \(pq\), we get the six terms \(a^2b,a^2c,b^2a,b^2c,c^2a,c^2b\) and three copies of \(abc\). Thus \(pq=\sum_{\mathrm{sym}}a^2b+3r\), which gives the formula.

The right side equals \(\frac12(2a^2+2b^2+2c^2-2ab-2bc-2ca)=a^2+b^2+c^2-q\). Also \(p^2-3q=(a^2+b^2+c^2+2q)-3q=a^2+b^2+c^2-q\).

The difference between the left and right sides is \((p^3-3pq+3r)+3r-(pq-3r)=p^3-4pq+9r\). Therefore the original inequality is equivalent to the nonnegativity of this expression.

The inequality is equivalent to \(p^2-2q\ge q\), i.e. \(p^2-3q\ge0\). But \(p^2-3q=\frac12((a-b)^2+(b-c)^2+(c-a)^2)\ge0\).

By \(p^2-3q=\frac12\sum(a-b)^2\ge0\), we get \(q\le\frac{p^2}{3}\). Equality is possible only at \(a=b=c\).

By AM-GM, \(\frac{a+b+c}{3}\ge\sqrt[3]{abc}\). Hence \(\sqrt[3]{abc}\le p/3\), so \(abc\le p^3/27\). Equality holds at \(a=b=c=p/3\).

The left side is \(t^3+2+3t\). The right side is \(2t^2+2t+2\). The difference is \(t^3-2t^2+t=t(t-1)^2\ge0\).

By Problem 5, the difference between the left and right sides is \(p^3-4pq+9r\). By Schur degree \(3\), this expression is nonnegative for \(a,b,c\ge0\). Hence the inequality follows.

We have \(\sum a^2=p^2-2q\). Also \(a^2b^2+b^2c^2+c^2a^2=q^2-2pr\). Thus \[\sum a^4=(p^2-2q)^2-2(q^2-2pr)=p^4-4p^2q+2q^2+4pr.\]

By the previous problem, \(\sum a^4=p^4-4p^2q+2q^2+4pr\). Subtracting \(abc(a+b+c)=pr\), we obtain \(p^4-4p^2q+2q^2+3pr\).

After substitution, we get \(t^4+2\ge t(t+2)\). The difference is \(t^4-t^2-2t+2=(t-1)^2(t^2+2t+2)\ge0\).

By UVW, it is enough to check the two-equal case and the boundary. If \(b=c=1\), \(a=t\), the difference is \((t-1)^2(t^2+2t+2)\ge0\). If one variable is \(0\), the right side is \(0\), while the left side is nonnegative. Hence the inequality holds.

The left side is \(t^4+2+t(t+2)=t^4+t^2+2t+2\). The right side is \(2t^3+2t+2\). The difference is \(t^4-2t^3+t^2=t^2(t-1)^2\ge0\).

The inequality is symmetric and of Schur type in 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, say \(c=0\), we get \(a^4+b^4\ge a^3b+ab^3\), i.e. \((a-b)^2(a^2+ab+b^2)\ge0\). Hence the inequality is proved.

Let \(x=ab\), \(y=bc\), \(z=ca\). Then \((x+y+z)^2\ge3(xy+yz+zx)\). The left side is \(q^2\), and \(xy+yz+zx=abc(a+b+c)=pr\). Therefore \(q^2\ge3pr\).

Since \(p=1\), \(\sum a^3=1-3q+3r\). The left side minus the right side equals \(1-4q+9r\). This is Schur degree \(3\) with \(p=1\): \(p^3-4pq+9r\ge0\). Hence the inequality holds.

By UVW, for fixed \(p\), the extremum of \(r=abc\) occurs when two variables are equal or on the boundary. On the boundary \(r=0\). Let \(b=c=x\), \(a=p-2x\). Then \[ \frac{p^3}{27}-(p-2x)x^2=\frac{(p-3x)^2(p+6x)}{27}\ge0. \] Hence \(abc\le p^3/27\).

The general Schur degree \(4\) is \(\sum a^4+abc(a+b+c)\ge\sum_{\mathrm{sym}}a^3b\). Under \(a+b+c=3\), we have \(abc(a+b+c)=3abc\). Thus the required inequality follows directly from Schur degree \(4\).

The first expression is multiplied by \(t^2\), so its degree is \(2\). The second does not change, so its degree is \(0\). The third is multiplied by \(t\), so its degree is \(1\).

The first is homogeneous of degree \(2\), so normalization is allowed. The second is not homogeneous: the degrees are \(2\) and \(0\), so normalization is not allowed. The third has degree \(0\), so normalization is allowed for positive variables.

The inequality is homogeneous. With \(a+b+c=1\), we need \(a^2+b^2+c^2\ge1/3\). This follows from \((a+b+c)^2\le3(a^2+b^2+c^2)\). Returning to the original scale gives the result.

Set \(x=a/\sqrt[3]{abc}\), \(y=b/\sqrt[3]{abc}\), \(z=c/\sqrt[3]{abc}\). Then \(xyz=1\). We need \(x+y+z\ge3\), which follows from AM-GM.

Since \(a+b+c=1\), it is enough to prove \(ab+bc+ca\le\frac{(a+b+c)^2}{3}\), i.e. \(3q\le p^2\). This follows from \(p^2-3q=\frac12\sum(a-b)^2\ge0\).

Let \(x=a/b\). The inequality becomes \(x+\frac1x\ge2\), which follows from \((x-1)^2\ge0\) after multiplying by \(x>0\).

The expression has degree \(0\). By AM-GM, \[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge3\sqrt[3]{\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{c}{a}}=3.\]

By AM-GM, \[\sum\frac{a^2}{b^2}\ge3\sqrt[3]{\frac{a^2}{b^2}\cdot\frac{b^2}{c^2}\cdot\frac{c^2}{a^2}}=3.\]

By Cauchy, \[\sum\frac{a^2}{a+b}\ge\frac{(a+b+c)^2}{(a+b)+(b+c)+(c+a)}=\frac{(a+b+c)^2}{2(a+b+c)}=\frac{a+b+c}{2}.\]

By Cauchy, \[\sum\frac{a}{b+c}=\sum\frac{a^2}{a(b+c)}\ge\frac{(a+b+c)^2}{2(ab+bc+ca)}.\] Since \((a+b+c)^2\ge3(ab+bc+ca)\), we get the lower bound \(\frac32\).

By AM-GM, \(\frac{a+b+c}{3}\ge\sqrt[3]{abc}\). Since the sum is \(1\), \(\sqrt[3]{abc}\le1/3\), hence \(abc\le1/27\). Homogeneous form: \(abc\le\frac{(a+b+c)^3}{27}\).

By AM-GM, \(\frac{a^2+b^2+c^2}{3}\ge\sqrt[3]{a^2b^2c^2}=1\). Therefore \(\sum a^2\ge3\).

By Cauchy, \((1+1+1)^2\le(a+b+c)\left(\frac1a+\frac1b+\frac1c\right)\). Therefore the product is at least \(9\).

By Cauchy, \((a+b+c)^2\le3(a^2+b^2+c^2)\). Since \(a+b+c=3\), we get \(9\le3\sum a^2\), so \(\sum a^2\ge3\).

By Cauchy, \[\sum\frac{a^2}{b+c}\ge\frac{(a+b+c)^2}{(b+c)+(c+a)+(a+b)}=\frac{(a+b+c)^2}{2(a+b+c)}=\frac{a+b+c}{2}.\]

If \(a+b+c=0\), the result is obvious. Otherwise set \(S=a+b+c\), \(x=a/S\), \(y=b/S\), \(z=c/S\). Then \(x+y+z=1\), so \(\frac{a^2+b^2+c^2}{S^2}\ge1/3\). Multiply by \(S^2\).

This is a symmetric homogeneous inequality of degree \(4\). By the UVW form already proved, it is enough to check \(b=c=1\), \(a=t\): \(t^4+2\ge t(t+2)\), equivalent to \((t-1)^2(t^2+2t+2)\ge0\). On the boundary, the right side is \(0\). Hence the inequality holds.

The expression has degree \(0\). Divide \(a,b,c\) by \(\sqrt[3]{abc}\); the product becomes \(1\), and the expression does not change. With \(abc=1\), we need \(a^2+b^2+c^2\ge3\), which follows from AM-GM.

Let \(q=ab+bc+ca\). Since \(p^2=\sum a^2+2q=1\), we have \(\sum a^2+q=1-q\). Also \(q\le1/3\), so \(1-q\ge2/3\). The homogeneous version is \[\sum a^2+ab+bc+ca\ge\frac{2}{3}(a+b+c)^2.\]

The expression has degree \(0\), so normalize \(abc=1\). Then \(\sqrt[3]{a^2b^2c^2}=1\). We get \[\frac{\sum a^2}{q}+q.\] Since \(\sum a^2\ge q\), the first fraction is at least \(1\). Also \(q=ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}=3\). Hence the whole sum is at least \(1+3=4\).

Set \(x=a/b\). Then \(b/a=1/x\), and \(x+\frac1x-2=\frac{(x-1)^2}{x}\ge0\).

Then \(xyz=1\). By AM-GM, \(x+y+z\ge3\sqrt[3]{xyz}=3\), as required.

Let \(a=x^2\), \(b=y^2\), \(x,y\ge0\). We need \(x+y\le\sqrt{2(x^2+y^2)}\). Squaring gives \(2xy\le x^2+y^2\), i.e. \((x-y)^2\ge0\).

Set \(x=(b+c-a)/2\), \(y=(c+a-b)/2\), \(z=(a+b-c)/2\). Triangle inequalities give \(x,y,z>0\). Direct checking gives \(y+z=a\), \(z+x=b\), \(x+y=c\).

After substitution, \(2(ab+bc+ca)-(a^2+b^2+c^2)=4(xy+yz+zx)>0\). Hence the strict inequality holds.

Let \(a=y+z\), \(b=z+x\), \(c=x+y\). The sum becomes \(\frac12\left(\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\right)\). Inside are the pairs \(\frac{y}{x}+\frac{x}{y}\), \(\frac{z}{x}+\frac{x}{z}\), \(\frac{z}{y}+\frac{y}{z}\), each at least \(2\). Thus the sum is at least \(3\).

We have \(x+y+z=0\). Hence \(\sum a^2=\sum(1+x)^2=3+2\sum x+\sum x^2=3+\sum x^2\ge3\).

For \(x,y\ge0\), there is \(t\in[0,\pi/2]\) such that \(x=\cos t\), \(y=\sin t\). Then \(xy=\frac12\sin2t\le\frac12\).

After substitution, we need \(xy+yz+zx\le x^2+y^2+z^2\). This is equivalent to \(\frac12((x-y)^2+(y-z)^2+(z-x)^2)\ge0\).

Set \(z=1\), \(y=b\), \(x=ab\). Then \(x/y=a\), \(y/z=b\), and \(z/x=1/(ab)=c\), since \(abc=1\).