3. Let $0<x<1, a, b$ be positive constants, then the minimum value of $\frac{a^{2}}{x}+\frac{b^{2}}{1-x}$ is $\qquad$
Solve:
$\frac{a^{2}}{x}+\frac{b^{2}}{1-x} \geqslant \frac{(a+b)^{2}}{x+1-x}=(a+b)^{2}$, equality holds when $x=\frac{a}{a+b}$. Therefore, the minimum value of $\frac{a^{2}}{x}+\frac{b^{2}}{1-x}$ is $(a+b)^{2}$.