Mock Olympiads: Inequalities

The left side is \(\frac{a^2+b^2+c^2}{ab+bc+ca}\). Since \(a^2+b^2+c^2\ge ab+bc+ca\), it is at least \(1\).

For \(x>0\), \(\frac{x}{x^2+x+1}\le\frac13\), because \(x^2-2x+1=(x-1)^2\ge0\). The sum of three such fractions is at most \(1\).

We have \(3x^3-(2x-y)(x^2+xy+y^2)=(x-y)^2(x+y)\ge0\). Summing cyclically gives \(\frac{2a-b+2b-c+2c-a}{3}=\frac{a+b+c}{3}\).

By Schur, \(\sum a^3(a-b)(a-c)\ge0\). Expanding gives exactly \(\sum a^5+abc\sum a^2-\sum_{\mathrm{sym}}a^4b\ge0\).

From \((a+b+c)^2\ge3(ab+bc+ca)\) and \(a+b+c=1\), we get \(1\ge3(ab+bc+ca)\).

The sum of denominators is \(2(a^2+b^2+c^2)+2(ab+bc+ca)=2(a+b+c)^2\). Thus by Cauchy the left side is at least \((a+b+c)^2/(2(a+b+c)^2)=1/2\).

Since \(x^2+xy+y^2-\frac34(x+y)^2=\frac14(x-y)^2\ge0\), \(\sqrt{x^2+xy+y^2}\ge\frac{\sqrt{3}}2(x+y)\). Summing gives \(3\sqrt{3}\).

We have \(A>0\), \(C=P(0)>0\), and \(B^2-4AC<0\). The quadratic \(Cx^2+Bx+A\) has leading coefficient \(C>0\) and the same discriminant, so it is positive for all \(x\).

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

Then the denominators are \(2x,2y,2z\), and the sum is \(\frac12\sum\frac{y+z}{x}\). It contains three pairs \(\frac{x}{y}+\frac{y}{x}\), each at least \(2\). Hence the sum is at least \(3\).

By Holder, the left side is at least \(\frac{(a+b+c)^3}{2(a^2+b^2+c^2)+ab+bc+ca}\). Since the denominator is at most \(3(a+b+c)^2\), the result follows.

Let \(y_1<\cdots

By AM-GM, \(a+b+c\ge3\). Also \(ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}=3\). Add the estimates.

By Cauchy, the left side is at least \(\frac{(a+b+c)^2}{a^2+b^2+c^2+3ab+3bc+3ca}\). Comparing this fraction with \(3/4\) is equivalent to \(a^2+b^2+c^2\ge ab+bc+ca\).

The difference \(\frac1{1+x^2}-\frac{2-x}{2}\) equals \(\frac{x(x-1)^2}{2(1+x^2)}\ge0\). Sum for \(a,b,c\) and use \(a+b+c=3\).

For \(u,v>0\), \((v-u)\sqrt{u^2+3v^2}\ge v^2-u^2\): if \(v\ge u\), the root is at least \(u+v\); if \(v\le u\), the root is at most \(u+v\), but the multiplier is negative. The right-hand sides telescope to \(0\).

By AM-GM, \(2a^4+b^4+c^4\ge4a^2bc\), and similarly for the other two mixed terms. Adding the three estimates gives \(4\sum a^4\ge4abc(a+b+c)\).

By Cauchy, \((\sum a^3)^2\le(\sum a^2)(\sum a^4)\). Also, \(\sum a^4\le(\sum a^2)^2\). Hence \((\sum a^3)^2\le(\sum a^2)^3\).

By Cauchy, \[\sum\frac1{a+2}\ge\frac9{a+b+c+6}=\frac9{9}=1.\]

Let \(y_1<\cdots