Prove that $\frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)}\ge \frac32$ 
$a,b,c$ 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.
 A: 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.
A: As above write $x=a^{-1}$, $y=b^{-1}$, $c=z^{-1}$, so we are required to prove
$$ \frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} \ge \frac{3}{2}, \qquad xyz=1.$$
Consider now that $\frac{x}{y+z}$, $\frac{y}{z+x}$, $\frac{z}{x+y}$ are in the same order as $x,y,z$. Then by the Rearrangement Inequality
$$ \frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} \ge \frac{xy}{y+z} + \frac{yz}{z+x} + \frac{zx}{x+y} $$
and also
$$ \frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} \ge \frac{xz}{y+z} + \frac{yz}{z+x} + \frac{zy}{x+y}; $$
averaging the two, we get
$$ \frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} \ge \frac{x+y+z}{2} \ge \frac{3}{2}$$
the last step forllowing from AM-GM.
A: $$ AM \ge GM $$
$$\frac{\frac{1}{a_{3}(b+c)} + \frac{1}{b_{3}(a+c)} + \frac{1}{c_{3}(a+b)}}{3} \ge \sqrt[3]{\frac{1}{(a+b)(b+c)(c+a)}}$$
$$  \frac{1}{a_{3}(b+c)} + \frac{1}{b_{3}(a+c)} + \frac{1}{c_{3}(a+b)} \ge \frac{3}{\sqrt[3]{2abc+ab^2 + ba^2 + ac^2 + ca^2 + bc^2 + cb^2}}$$
$$AM \ge GM $$
$$  \frac{2abc+ab^2 + ba^2 + ac^2 + ca^2 + bc^2 + cb^2}{8} \ge \sqrt[8]{a^8b^8c^8}$$
$$ \ge \frac{3}{\sqrt[3]{8\sqrt[8]{a^8b^8c^8}}}$$
$$ \ge \frac{3}{\sqrt[3]{8}}$$
$$ \ge \frac{3}{2}$$
