$\frac{a}{a+2b+c}+\frac{b}{b+2c+a}+\frac{c}{c+2a+b}\geq\frac{3}{4}$

Question

I found the following exercise: Prove that $$\frac{a}{a+2b+c}+\frac{b}{b+2c+a}+\frac{c}{c+2a+b}\geq\frac{3}{4}$$ for any positive $a$, $b$, $c$.

I tried substituting the denominators but it led me nowhere. I don really have other ideas.

Answer 1

Using Cauchy-Schwarz:

$$ \left(\sum_{cyc}\frac{a^2}{a^2+2ab+ca}\right)\left[\sum_{cyc}(a^2+2ab+ca)\right] \geq \left(\sum_{cyc}\frac{a}{\sqrt{a^2+2ab+ca}}\cdot \sqrt{a^2+2ab+ca}\right)^2$$

$$=(a+b+c)^2$$

Since $\displaystyle\sum_{cyc}(a^2+2ab+ca) =a^2+b^2+c^2+3(ab+bc+ca)$, we get:

$$\sum_{cyc}\frac{a^2}{a^2+2ab+ca}\geq \frac{(a+b+c)^2}{a^2+b^2+c^2+3(ab+bc+ca)}$$

Therefore

$$\sum_{cyc} \frac{a}{a+2b+c}=\sum_{cyc}\frac{a^2}{a^2+2ab+ca}\geq \frac{(a+b+c)^2}{a^2+b^2+c^2+3(ab+bc+ca)}$$

It remains to prove:

$$\frac{(a+b+c)^2}{a^2+b^2+c^2+3(ab+bc+ca)} \geq \frac{3}{4}$$

or after expanding

$$a^2+b^2+c^2\geq ab+bc+ca$$

which is

$$\frac{1}{2}\left[(a-b)^2+(b-c)^2+(c-a)^2\right]\geq 0$$

Answer 2

A full expanding gives $$\sum_{cyc}(2a^3-a^2b+3a^2c-4abc)\geq0$$ or $$\sum_{cyc}(a^3-a^2b)+\sum_{cyc}(a^3+3a^2c-4abc)\geq0,$$ which is true by Rearrangement and AM-GM.