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Problem 814

AMC 12 late, AIME early

Algebra Difficulty 4.5 Prove it Indian National Mathematical Olympiad · India

If $a, b, c$ are positive real numbers such that $a b c = 1$ , prove that
$a^{b+c} b^{c+a} c^{a+b} \leq 1$

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

Solution:
Note that the inequality is symmetric in $a, b, c$ so that we may assume that $a \geq b \geq c$ . Since $a b c = 1$ , it follows that $a \geq 1$ and $c \leq 1$ . Using $b = 1 / (a c)$ , we get
$a^{b+c} b^{c+a} c^{a+b} = \frac{a^{b+c} c^{a+b}}{a^{c+a} c^{c+a}} = \frac{c^{b-c}}{a^{a-b}} \leq 1$
because $c \leq 1$ , $b \geq c$ , $a \geq 1$ and $a \geq b$ .

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.