Prove that $\frac{1}{a^{3}(b+c)} +\frac{1}{b^{3}(a+c)} +\frac{1}{c^{3}(a+b)} \ge \frac{3}{2}$

Question

Prove that $\frac{1}{a^{3}(b+c)} +\frac{1}{b^{3}(a+c)} +\frac{1}{c^{3}(a+b)} \ge \frac{3}{2}$ if abc=1

My approach A= $\frac{1}{a^{3}(b+c)}$,B= $\frac{1}{b^{3}(c+a)}$ &C=$\frac{1}{c^{3}(a+b)}$

A.M$\ge$G.M

$\frac{A+B+C}{3} \ge ({ABC})^{\frac{1}{3}}$

As abc=1

$\frac{A+B+C}{3} \ge ({\frac{1}{(b+c)(a+c)(a+b)}})^\frac{1}{3}$

I am struck after this step, I even tried to compare G.M$\ge$H.M but not able to eliminate the terms

Answer 1

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