Track / Stage 7 / 124 of 300 #1524 of 1964

Problem 1524

National olympiad second round; IMO P1/P4

Algebra Difficulty 7.3 Prove it

Theorem 2 If $a, b, c \in R^{+}$ , then
$\frac{a b}{a^{2}+b^{2}+2 c^{2}}+\frac{b c}{2 a^{2}+b^{2}+c^{2}}+\frac{c a}{a^{2}+2 b^{2}+c^{2}} \leq \frac{3}{4} \text {. }$

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Theorem 2 Proof: Applying the 2-variable mean inequality, we get
$\begin{array}{l} \frac{a b}{a^{2}+b^{2}+2 c^{2}}+\frac{b c}{2 a^{2}+b^{2}+c^{2}}+\frac{c a}{a^{2}+2 b^{2}+c^{2}}= \\ \frac{a b}{\left(b^{2}+c^{2}\right)+\left(c^{2}+a^{2}\right)}+\frac{b c}{\left(c^{2}+a^{2}\right)+\left(a^{2}+b^{2}\right)}+\frac{c a}{\left(a^{2}+b^{2}\right)+\left(b^{2}+c^{2}\right)} \leq \\ \frac{a b}{2 \sqrt{\left(b^{2}+c^{2}\right)\left(c^{2}+a^{2}\right)}}+\frac{b c}{2 \sqrt{\left(c^{2}+a^{2}\right)\left(a^{2}+b^{2}\right)}}+\frac{a b}{2 \sqrt{\left(b^{2}+c^{2}\right)\left(c^{2}+a^{2}\right)}} \frac{1}{4} \\ \left(2 \sqrt{\frac{b^{2}}{b^{2}+c^{2}} \cdot \frac{a^{2}}{c^{2}+a^{2}}}+2 \sqrt{\frac{c^{2}}{c^{2}+a^{2}} \cdot \frac{b^{2}}{a^{2}+b^{2}}}+2 \sqrt{\frac{a^{2}}{a^{2}+b^{2}} \cdot \frac{c^{2}}{b^{2}+c^{2}}}\right) \leq \\ \frac{1}{4}\left(\frac{b^{2}}{b^{2}+c^{2}}+\frac{a^{2}}{c^{2}+a^{2}}+\frac{c^{2}}{c^{2}+a^{2}}+\frac{b^{2}}{a^{2}+b^{2}}+\frac{a^{2}}{a^{2}+b^{2}}+\frac{c^{2}}{b^{2}+c^{2}}\right)=\frac{1}{4} \text { . } \end{array}$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.