Track / Stage 7 / 36 of 300 #1436 of 2000

Problem 1436

National olympiad second round; IMO P1/P4

Algebra Difficulty 7.1 Prove it

$\begin{aligned} & \text { Given } a>0, b>0 \text {, prove that } \\ & \sqrt{\frac{a}{2 a+b}}+\sqrt{\frac{b}{a+2 b}} \leqslant \sqrt{\frac{a}{a+2 b}}+\sqrt{\frac{b}{2 a+b}} \\ \leqslant & \frac{2}{\sqrt{3}} . \end{aligned}$

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Proof: First, prove the left inequality, eliminate the denominator
$\begin{array}{l} \Leftrightarrow \sqrt{a(a+2 b)}+\sqrt{b(2 a+b)} \leqslant \sqrt{a(2 a+b)} \\ +\sqrt{b(a+2 b)} \\ \Leftrightarrow a(a+2 b)+b(2 a+b) \leqslant a(2 a+b)+b(a+ \end{array}$
$2 b)$
$\Leftrightarrow 2 a b \leqslant a^{2}+b^{2} .$

Next, prove the right inequality, eliminate the denominator
$\begin{aligned} \Leftrightarrow & \sqrt{3 a(2 a+b)}+\sqrt{3 b(a+2 b)} \\ \leqslant & 2 \sqrt{(2 a+b)(a+2 b)} \\ \Leftrightarrow & 3 a(2 a+b)+3 b(a+2 b) \\ & +6 \sqrt{a b(2 a+b)(a+2 b)} \\ \leqslant & 4(2 a+b)(a+2 b) \\ \Leftrightarrow & 3 \sqrt{a b(2 a+b)(a+2 b)} \leqslant a^{2}+b^{2}+7 a b \\ \Leftrightarrow & 9(a+2 b)(2 a+b) \leqslant\left(a^{2}+b^{2}+7 a b\right)^{2} \\ \Leftrightarrow & 4 a^{3} b+4 a b^{3} \leqslant a^{4}+b^{4}+6 a^{2} b^{2} \\ \Leftrightarrow & (a-b)^{2} \geqslant 0 . \end{aligned}$

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