For simplicity, let's write $g(x) = \frac{a}{x^2}$ and $h(x)=\frac{b}{1-x}$. Then $f=\max\lbrace g,h\rbrace$. Note that $g$ is strictly decreasing and $h$ is strictly increasing. The function $g-f$ is a continuous function such that $\lim\limits_{x\downarrow 0}=+\infty$ and $\lim\limits_{x\uparrow1}=-\infty$. Since $g-f$ is continuous, it therefore must have a zero somewhere on $(0,1)$. (Added: this zero is unique, since $g-f$ is strictly decreasing, as it is the sum of stricly decreasing functions $g$ and $-f$.)
So the two graphs will necessarily intersect in a unique point $x_0\in(0,1)$. Now, $f(x)=g(x)$ for $x\in(0,x_0]$ and $f(x)=h(x)$ for $x\in[x_0,1)$, so $f$ is strictly decreasing on $(0,x_0]$ and strictly increasing on $[x_0,1)$. This implies the only possible point where it might achieve a minimum is $x_0$ and also that it indeed achieves a minimum there. To calculate the point $x_0$, just solve the quadratic equation you get.