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

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.

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.