P.T. $\frac{1}{a^3(b+c)} +\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)} \ge \frac 32$ 
If $abc=1$ where $a,b,c$ are positive real. Prove that ,$\frac{1}{a^3(b+c)} +\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)} \ge \frac 32$.

I tried to multiply the LHS by $abc$ to make the relation homogeneous but......
There is a $3$ in the RHS ,So I applied AM-GM and found that it is enough to show that 
$(a+b)(b+c)(c+a) \le 8$ where $abc=1$.
I know it is a very well-known inequality but I couldn't answer it . Please help me.

Source: It is derived from a question came in any Olympiad question (I cannot remember, now).
It is a good problem for the application of AM-GM or, Cauchy-Schartz.Jensen's inequality can also help.

By C-S and AM-GM we obtain:
$$\sum_{cyc}\frac{1}{a^3(b+c)}=\sum_{cyc}\frac{b^2c^2}{a(b+c)}\geq\frac{(ab+ac+bc)^2}{\sum\limits_{cyc}a(b+c)}=$$
$$=\frac{(ab+ac+bc)^2}{2(ab+ac+bc)}=\frac{1}{2}(ab+ac+bc)\geq\frac{3}{2}\sqrt[3]{a^2b^2c^2}=\frac{3}{2}.$$
