Help with inequality problem

Question

Given $a$ , $b$ , $c \ge 0$ show that $$\frac{a^2}{(a+b)(a+c)} + \frac{b^2}{(b+a)(b+c)}+ \frac{c^2}{(c+a)(c+b)} \ge \frac{3}{4}.$$

I tried using Titu's lemma on it, resulting in

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

And I am stuck here.

Answer 1

By C-S $$\sum_{cyc}\frac{a^2}{(a+b)(a+c)}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(a+b)(a+c)}\geq\frac{3}{4},$$ where the last inequality it's $$4\sum_{cyc}(a^2+2ab)\geq3\sum_{cyc}\left(a^2+3ab\right)$$ or $$\sum_{cyc}(a^2-ab)\geq0$$ or $$\sum_{cyc}(2a^2-2ab)\geq0$$ or $$\sum_{cyc}(a^2+b^2-2ab)\geq0$$ or $$\sum_{cyc}(a-b)^2\geq0.$$

Answer 2

Another way.

We need to prove that $$4\sum_{cyc}(a^2b+a^2c)\geq3\prod_{cyc}(a+b)$$ or $$\sum_{cyc}c(a-b)^2\geq0.$$

Answer 3

Hint.

As the inequality is homogeneous, making the substitutions $b=\lambda a, c=\mu a$ we have

$$\frac{a^2}{(a+b)(a+c)} + \frac{b^2}{(b+a)(b+c)}+ \frac{c^2}{(c+a)(c+b)} = \frac{\lambda ^2 \mu +\lambda ^2+\lambda \mu ^2+\lambda +\mu ^2+\mu }{(\lambda +1) (\mu +1) (\lambda +\mu )} = f(\lambda,\mu) $$ with stationary points as the solutions $(\lambda^*,\mu^*)$ for

$$ \nabla f = 0\Rightarrow \cases{\lambda ^2 \mu -\mu ^2 = 0\\ \lambda ^2-\lambda \mu ^2=0} $$

now discarding the solution $\lambda=0,\mu = 0$ we have to demonstrate that $f(\lambda^*,\mu^*)\ge \frac 34$.