Proving a small inequality I am given that $a^3+b^3+c^3=3$ where a, b, c are positive numbers and I need to prove that
$$\frac {3(ab+bc+ac)+a^3c^2+b^3a^2+c^3b^2}{(a+b)(b+c)(a+c)}\ge \frac {3}{2}$$
At first it seems to me that the inequality might be wrong. I have tried using the Cauchy Schwarz , AM GM and some algebraic manipulations to reach the inequality but none of them helped me out. 
Can somebody around  here help me to prove this small inequality. Thanks in advance.

We need to prove that
$$\sum_{cyc}\frac{3ab+a^3c^2}{\prod\limits_{cyc}(a+b)}\geq\frac{3}{2}$$ or
$$\sum_{cyc}\frac{(a^3+b^3+c^3)ab+a^3c^2}{\prod\limits_{cyc}(a+b)}\geq\frac{3}{2}$$ or
$$\sum_{cyc}\frac{a^4b+a^4c+a^3bc+a^3c^2}{\prod\limits_{cyc}(a+b)}\geq\frac{3}{2}$$ or
$$\sum_{cyc}\frac{a^3(ab+ac+bc+c^2)}{\prod\limits_{cyc}(a+b)}\geq\frac{3}{2}$$ or
$$\sum_{cyc}\frac{a^3(b+c)(c+a)}{\prod\limits_{cyc}(a+b)}\geq\frac{3}{2}$$or
$$\sum_{cyc}\frac{a^3}{a+b}\geq\frac{3}{2}$$ and see here:
If $a^3+b^3+c^3=3$ so $\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a}\geq\frac{3}{2}$
