For any $\displaystyle a, b, c > 0$ with $\displaystyle abc = 1$, prove that
$\displaystyle \frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3 (a+b)} \ge \frac{3}{2}$
$\displaystyle a(b+c)+b(a+c)+c(a+b)=2(ab+bc+ac)=2\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right),$ and $\displaystyle ab+bc+ac\ge 3\sqrt[3]{a^{2}b^{2}c^{2}}=3,$ now by Cauchy - Schwarz we have $\displaystyle
\left( \frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(a+c)}+\frac{1} {c^{3}(a+b)} \right)2(ab+bc+ac)\ge \left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)^{2},$ finally $\displaystyle \left( \frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(a+c)}+\frac{1} {c^{3}(a+b)} \right)\ge \frac{ab+bc+ac}{2}\ge \frac{3}{2}.\quad\blacksquare$