Track / Stage 7 / 16 of 300 #1416 of 2000

Problem 1416

National olympiad second round; IMO P1/P4

Algebra Difficulty 7.0 Prove it

78. Let $a, b, c$ be positive real numbers, and $a b c=1$ , prove that: $\frac{1}{1+2 a}+\frac{1}{1+2 b}+\frac{1}{1+2 c} \geqslant 1$ .
(2004 German IMO Team Selection Exam Problem)

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

78. By the AM-GM inequality,

So $\quad \frac{1}{1+2 a} \geqslant \frac{1}{1+\left(\frac{a}{b}\right)^{\frac{2}{3}}+\left(\frac{a}{c}\right)^{\frac{2}{3}}}=\frac{(b c)^{\frac{2}{3}}}{(a b)^{\frac{2}{3}}+(b c)^{\frac{2}{3}}+(c a)^{\frac{2}{3}}}$
$\begin{array}{l} \left(\frac{a}{b}\right)^{\frac{2}{3}}+\left(\frac{a}{c}\right)^{\frac{2}{3}} \geqslant 2 \sqrt{\left(\frac{a}{b}\right)^{\frac{2}{3}} \cdot\left(\frac{a}{c}\right)^{\frac{2}{3}}}= \\ 2 \sqrt{\left(\frac{a^{2}}{b c}\right)^{\frac{2}{3}}}=2 \sqrt{\left(\frac{a^{2}}{1 / a}\right)^{\frac{2}{3}}}=2 a \end{array}$
$\frac{1}{1+2 b} \geqslant \frac{(-c a)^{\frac{2}{3}}}{(a b)^{\frac{2}{3}}+(b c)^{\frac{2}{2}}+(c a)^{\frac{2}{3}}}, \frac{1}{1+2 c} \geqslant \frac{(a b)^{\frac{2}{3}}}{(a b)^{\frac{2}{3}}+(b c)^{\frac{2}{2}}+(c a)^{\frac{2}{3}}}$

Adding the above three inequalities, we get $\frac{1}{1+2 a}+\frac{1}{1+2 b}+\frac{1}{1+2 c} \geqslant 1$ .

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