IMO 1995 - Problem 2

Problem: Let $a,b$ and $c$ be positive real numbers such that $abc=1$. Prove that: $\frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)}\ge \frac{3}{2}$.
Solution: Let $S$ be the right hand side and $T=a(b+c)+b(a+c)+c(a+b)=2(ab+bc+ca)\ge 6(abc)^{\frac{2}{3}}=6$. Then by Cauchy's inequality, we have :
$ST\ge (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2=\frac{T^2}{4}$.
Thus $S\ge \frac{T}{4}\ge \frac{3}{2}$ as desired.
Second solution: Let $x=\frac{1}{a},y=\frac{1}{b},z=\frac{1}{c}$. Then $xyz=1$ and $S=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}$.
By Cauchy's inequality,
[(x+y)+(z+x)+(x+y)] S\ge (x+y+z)^2.
or $S\ge \frac{x+y+z}{2}\ge \frac{3}{2}(xyz)^{\frac{1}{3}}=\frac{3}{2}$ as desired.
Source: https://sms.math.nus.edu.sg/Simo/IMO_Problems/95.pdf