The minimum value of the function $f(x)=\frac{a^2}{x}+\frac{b^2}{a-x}$, a > 0, b > 0, in (0, a), is

$\frac{(a+b)^2}{a}$

We have,

$f(x)=\frac{a^2}{x}+\frac{b^2}{a-x}$

$\Rightarrow f'(x)=-\frac{a^2}{x^2}+\frac{b^2}{(a-x)^2}$

∴  $f'(x)=0$

$\Rightarrow \frac{a^2}{x^2}=\frac{b^2}{(a-x)^2}$

$\Rightarrow a^2(a-x)^2=b^2 x^2$

$\Rightarrow \ a(a-x)= \pm b x$

$\Rightarrow \ x=\frac{a^2}{a \pm b} \Rightarrow x=\frac{a^2}{a+b}$             $\left[∵ \frac{a^2}{a-b} \notin(0, a)\right]$

Now, $f''(x)=\frac{2 a^2}{x^3}+\frac{2 b^2}{(a-x)^2}>0$ for all $x \in(0, a)$

Hence, f(x) is minimum at $x=\frac{a^2}{a+b}$ with the minimum value given by

$f\left(\frac{a^2}{a+b}\right)=(a+b)+\frac{b^2}{a-\frac{a^2}{a+b}}=(a+b)+\frac{b^2(a+b)}{a b}=\frac{(a+b)^2}{a}$