Prove: $\frac{b+c}{a^2+bc}+\frac{c+a}{b^2+ac}+\frac{a+b}{c^2+ab}\leq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$

Question

Let $a;b;c>0$. Prove that :

$\frac{b+c}{a^2+bc}+\frac{c+a}{b^2+ac}+\frac{a+b}{c^2+ab}\leq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$

I think:

$\frac{b+c}{a^2+bc}+\frac{c+a}{b^2+ac}+\frac{a+b}{c^2+ab}\leq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\Leftrightarrow \frac{1}{a}-\frac{b+c}{a^2+ab}+\frac{1}{b}-\frac{c+a}{b^2+ac}+\frac{1}{c}-\frac{a+b}{c^2+ab}\geq 0\Leftrightarrow \frac{(a-b)(a-c)}{a^3+abc}+\frac{(b-c)(b-a)}{b^3+abc}+\frac{(c-a)(c-b)}{c^3+abc}\geq 0$

And then i don't know what i should do next ! :(

score 3 · Accepted Answer · answered Feb 17 '15 at 05:22

3

WLOG,you can assume $a\ge b\ge c>0$, then $$\frac{(b-c)(a-b)}{b^3+abc}\le\frac{(a-c)(b-c)}{c^3+abc}$$

answered Feb 17 '15 at 05:22

Nirvanacs

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Why can't I recognize it ? :(. Btw, Thanks very much :) – Lê Tấn Khang Feb 17 '15 at 05:24
$a-b\ge b-c$ isn't generally true, so how do you get the inequality? – Redundant Aunt Feb 17 '15 at 06:59

Prove: $\frac{b+c}{a^2+bc}+\frac{c+a}{b^2+ac}+\frac{a+b}{c^2+ab}\leq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$

1 Answers1