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Problem 1548

National olympiad second round; IMO P1/P4

Algebra Difficulty 7.3 Prove it

Given $a>0, b>0, 0<\lambda \leqslant \frac{1}{2}$ , prove
$\left(a^{\lambda}+b^{\lambda}\right) \cdot\left[\frac{1}{(2 a+b)^{\lambda}}+\frac{1}{(a+2 b)^{\lambda}}\right] \leqslant \frac{4}{3^{\lambda}} \text {. }$

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: Given $a>0, b>0, 0<2 \lambda \leqslant 1$ , using the power mean inequality, we get
$\begin{array}{l} \left(\frac{a}{2 a+b}\right)^{\lambda}+\left(\frac{b}{a+2 b}\right)^{\lambda} \\ =\left(\sqrt{\frac{a}{2 a+b}}\right)^{2 \lambda}+\left(\sqrt{\frac{b}{a+2 b}}\right)^{2 \lambda} \\ \leqslant 2^{1-2 \lambda}\left(\sqrt{\frac{a}{2 a+b}}+\sqrt{\frac{b}{a+2 b}}\right)^{2 \lambda} \\ \leqslant 2^{1-2 \lambda} \cdot\left(\frac{2}{\sqrt{3}}\right)^{2 \lambda} \\ =\frac{2}{3^{\lambda}} \end{array}$

Thus, we have $\left(\frac{a}{2 a+b}\right)^{\lambda}+\left(\frac{b}{a+2 b}\right)^{\lambda} \leqslant \frac{2}{3^{\lambda}}$ ,
Similarly, $\left(\frac{a}{a+2 b}\right)^{\lambda}+\left(\frac{b}{2 a+b}\right)^{\lambda} \leqslant \frac{2}{3^{\lambda}}$ .
Therefore, by adding the above two inequalities and transforming, we immediately obtain the inequality (5).

In particular, in inequality (5), taking $\lambda=\frac{1}{2}$ , we get inequality (2); taking $\lambda=\frac{1}{2}, a+b=1$ , we get inequality (1).

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