An inequality by substitution

AM-GM. $\frac1a+\frac1b=\frac{a+b}{ab}=\frac{1}{ab}$. By AM-GM, $ab\le\left(\frac{a+b}{2}\right)^2=\frac14$. So $\frac{1}{ab}\ge4$.

Cauchy-Schwarz. $(\sqrt{a}\cdot\frac{1}{\sqrt{a}}+\sqrt{b}\cdot\frac{1}{\sqrt{b}})^2\le(a+b)\left(\frac1a+\frac1b\right)=1\cdot\left(\frac1a+\frac1b\right)$. The LHS is $4$, so $4\le\frac1a+\frac1b$.

Conclusion. The inequality holds, with equality iff $a=b=1/2$.