Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 6.3 National olympiad Prove it

24 In $\triangle A B C$ , the three side lengths are $a, b, c$ , and $a, b, c$ are rational numbers. Prove:
$\left(1+\frac{b-c}{a}\right)^{a}\left(1+\frac{c-a}{b}\right)^{b}\left(1+\frac{a-b}{c}\right)^{c} \leqslant 1$

Solution

24. Since $a, b, c$ are positive rational numbers, there exists $m \in \mathbf{N}$ such that $m a, m b, m c$ are positive integers. Also, since $a, b, c$ are the lengths of the sides of a triangle, we have $1+\frac{b-c}{a}>0, 1+\frac{c-a}{b}>0, 1+\frac{a-b}{c}>0$ . By the AM-GM inequality, we get $\left[\left(1+\frac{b-c}{a}\right)^{m a}\left(1+\frac{c-a}{b}\right)^{m b}\left(1+\frac{a-b}{c}\right)^{m c}\right]^{\frac{1}{m a m b+m c}}$ $\frac{m a\left(1+\frac{b-c}{a}\right)+m b\left(1+\frac{c-a}{b}\right)+m c\left(1+\frac{a-b}{c}\right)}{m a+m b+m c}=1$ .

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