Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 6.1 National olympiad Prove it

Example 1.1.1. Let $a, b, c$ be three positive real numbers. Prove that
$a+b+c \leq \frac{a^{2}+b c}{b+c}+\frac{b^{2}+c a}{c+a}+\frac{c^{2}+a b}{a+b}$

Solution

SOLUTION. According to the identity
$\frac{a^{2}+b c}{b+c}-a=\frac{(a-b)(a-c)}{b+c}$
we can transform our inequality into the form
$x(a-b)(a-c)+y(b-a)(b-c)+z(c-a)(c-b) \geq 0$
where
$x=\frac{1}{b+c} ; y=\frac{1}{c+a} ; z=\frac{1}{a+b}$

WLOG, assume that $a \geq b \geq c$ , then clearly $x \leq y \leq z$ . The conclusion follows from the generalized Schur inequality instantly.

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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.