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. 

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