Prove that $\frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)}\ge \frac32$

Question

$a,b,c$ are positive reals with $abc = 1$. Prove that $$\frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)}\ge \frac32$$

I try to use AM $\ge$ HM.

$$\frac{\dfrac{1}{a^3(b+c)}+\dfrac{1}{b^3(a+c)}+\dfrac{1}{c^3(a+b)}}3\ge \frac{3}{a^3(b+c)+b^3(a+c)+c^3(a+b)}$$ Then how I proceed.

Answer 1

Hint: Let $x=a^{-1},y=b^{-1},z=c^{-1}$. Rewrite $a^3=\dfrac{a^2}{bc}$, the inequality becomes $$\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\ge\frac32,$$ where $xyz=1$. That should be easy by Cauchy-Schwarz.

Answer 2

$$ AM \ge GM $$

$$\frac{\frac{1}{a_{3}(b+c)} + \frac{1}{b_{3}(a+c)} + \frac{1}{c_{3}(a+b)}}{3} \ge \sqrt[3]{\frac{1}{(a+b)(b+c)(c+a)}}$$

$$ \frac{1}{a_{3}(b+c)} + \frac{1}{b_{3}(a+c)} + \frac{1}{c_{3}(a+b)} \ge \frac{3}{\sqrt[3]{2abc+ab^2 + ba^2 + ac^2 + ca^2 + bc^2 + cb^2}}$$

$$AM \ge GM $$ $$ \frac{2abc+ab^2 + ba^2 + ac^2 + ca^2 + bc^2 + cb^2}{8} \ge \sqrt[8]{a^8b^8c^8}$$

$$ \ge \frac{3}{\sqrt[3]{8\sqrt[8]{a^8b^8c^8}}}$$

$$ \ge \frac{3}{\sqrt[3]{8}}$$

$$ \ge \frac{3}{2}$$