A classic with abc = 1 (IMO 1995)

Substitution. With $x=\frac1a$ etc., $xyz=1$ and
$$\frac1{a^3(b+c)}=\frac{x^3}{\frac1y+\frac1z}=\frac{x^3yz}{y+z}=\frac{x^2}{y+z}.$$

Titu.
$$\sum\frac{x^2}{y+z}\ge\frac{(x+y+z)^2}{2(x+y+z)}=\frac{x+y+z}2.$$

AM-GM. $x+y+z\ge3\sqrt[3]{xyz}=3$, so the left side is at least $\frac32$.

Equality iff $a=b=c=1$.

Remark. The condition $abc=1$ was used twice: to simplify $x^3yz$ and to bound $x+y+z$.

Lesson. With $abc=1$, try $a=\frac1x$ (or $a=\frac yx$, $b=\frac zy$, $c=\frac xz$): it turns the condition into something usable and often reveals a known inequality.