Prove the inequality ${1 \over a^3 (b + c)} + {1 \over b^3 (a+c)} + {1 \over c^3 (a+b)} \ge \frac 32 $ given that $abc = 1$

Question

Prove the inequality $${1 \over a^3 (b + c)} + {1 \over b^3 (a+c)} + {1 \over c^3 (a+b)} \ge \frac 32 $$ So, I know a proof for this, but I basically memorized it without understanding. It's $$ {(\frac 1a + \frac 1b + \frac 1c)}^2 + {(a(b+c) + b(a+c) + c(a+b))}^-1 = {ab + bc + ac \over 2} \ge {\frac 32} $$ I don't know what inequalities were applied here or how. Another proof would also be appreciated.

Answer 1

we have $$\frac{1}{a^3(b+c)}=\frac{bc}{a^2(b+c)}=\frac{(bc)^2}{a(b+c)}$$ then we get $$\frac{(bc)^2}{a(b+c)}+\frac{(ac)^2}{b(a+c)}+\frac{(ab)^2}{c(a+b)}\geq \frac{(ab+bc+ac)^2}{2(ab+bc+ac)}\geq \frac{3}{2}$$ by $AM-GM$

Answer 2

By Holder and AM-GM $$\sum\limits_{cyc}\frac{1}{a^3(b+c)}=\sum_{cyc}\frac{b^3c^3}{b+c}\geq\frac{(ab+ac+bc)^3}{3\sum\limits_{cyc}(b+c)}\geq\frac{3abc(a+b+c)(ab+ac+bc)}{3\sum\limits_{cyc}(b+c)}\geq\frac{3}{2}$$