Minimising a/x + b/(1 − x)

Critical point. $f'(x)=-\dfrac a{x^2}+\dfrac b{(1-x)^2}=0\iff\dfrac{1-x}x=\sqrt{\dfrac ba}$ (both sides positive on $(0,1)$). So $x_0=\dfrac{\sqrt a}{\sqrt a+\sqrt b}$ and $1-x_0=\dfrac{\sqrt b}{\sqrt a+\sqrt b}$.

Nature. $f''(x)=\frac{2a}{x^3}+\frac{2b}{(1-x)^3}>0$: $f$ is convex, and $f\to+\infty$ at both ends, so $x_0$ is the global minimum.

Value.
$$f(x_0)=\frac{a(\sqrt a+\sqrt b)}{\sqrt a}+\frac{b(\sqrt a+\sqrt b)}{\sqrt b}=(\sqrt a+\sqrt b)(\sqrt a+\sqrt b)=(\sqrt a+\sqrt b)^2.$$

Consequence. For all $x\in(0,1)$ (and $y=1-x$): $\frac ax+\frac by\ge(\sqrt a+\sqrt b)^2$, with equality iff $x=x_0$.

Remark. For $a=b=1$: $\frac1x+\frac1y\ge4$ when $x+y=1$, the classical AM-HM inequality. The general inequality is a form of the Cauchy–Schwarz inequality: $\left(\frac ax+\frac by\right)(x+y)\ge(\sqrt a+\sqrt b)^2$.