2. Let $a>b>0$, then the minimum value of $a^{2}+\frac{1}{b(a-b)}$ is

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2. 4 Detailed Explanation: Since $a>b>0$, then $b(a-b) \leqslant\left[\frac{b+(a-b)}{2}\right]^{2}=\frac{a^{2}}{4}$, so $a^{2}+\frac{1}{b(a-b)}$ $\geqslant a^{2}+\frac{4}{a^{2}} \geqslant 4$, equality holds if and only if $a=\sqrt{2}, b=\frac{\sqrt{2}}{2}$.