contest math - Help with inequality problem

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

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 $$\sum_{cyc}(a-b)^2\geq0.$$