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$a,b,c$ 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.

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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.

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