Analysis: The original inequality is equivalent to $\frac{b^{2} c^{2}}{a(b+c)}+\frac{c^{2} a^{2}}{b(c+a)}$ $+\frac{a^{2} b^{2}}{c(a+b)} \geqslant \frac{3}{2}$. When $a=b=c=1$, equality holds, at which point $\frac{b^{2} c^{2}}{a(b+c)}=\frac{a(b+c)}{4}, \frac{c^{2} a^{2}}{b(c+a)}=\frac{b(c+a)}{4}, \frac{a^{2} b^{2}}{c(a+b)}$ $=\frac{c(a+b)}{4}$, and $\frac{b^{2} c^{2}}{a(b+c)}+\frac{a(b+c)}{4} \geqslant b c, \frac{c^{2} a^{2}}{b(c+a)}+$ $\frac{b(c+a)}{4} \geqslant c a, \frac{a^{2} b^{2}}{c(a+b)}+\frac{c(a+b)}{4} \geqslant a b$. Adding the last three inequalities side by side yields $\frac{b^{2} c^{2}}{a(b+c)}+\frac{c^{2} a^{2}}{b(c+a)}+\frac{a^{2} b^{2}}{c(a+b)} \geqslant \frac{1}{2}(a b$ $+b c+c a) \geqslant \frac{3}{2} \sqrt[3]{(a b c)^{2}}=\frac{3}{2}$.