Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 6.7 National olympiad Prove it

Example 3.2.3. Let $a, b, c$ be positive real numbers and $0 \leq k \leq 2$ . Prove that
$\frac{a^{2}-b c}{b^{2}+c^{2}+k a^{2}}+\frac{b^{2}-c a}{c^{2}+a^{2}+k b^{2}}+\frac{c^{2}-a b}{a^{2}+b^{2}+k c^{2}} \geq 0$

Solution

Solution. Although this problem can be solved in the same way as example 2.1.1 is solved, we can use Chebyshev inequality to give a simpler solution. Notice that if $a \geq b$ then for all positive real $c$ , we have $\left(a^{2}-b c\right)(b+c) \geq\left(b^{2}-c a\right)(c+a)$ , and
$\left(b^{2}+c^{2}+k a^{2}\right)(b+c)-\left(c^{2}+a^{2}+k b^{2}\right)(c+a)=(b-a)\left(\sum_{c y c} a^{2}-(k-1) \sum_{c y c} b c\right) \leq 0$

Having these results, we will rewrite the inequality into the following form
$\sum_{c y c} \frac{\left(a^{2}-b c\right)(b+c)}{(b+c)\left(b^{2}+c^{2}+k a^{2}\right)} \geq 0$
which is obvious by Chebyshev inequality because $\sum_{\text {cyc }}\left(a^{2}-b c\right)(b+c)=0$ .

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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.