Example 2 (2007 Guangzhou First Mock Exam for Science) Given $a, b>0$, and $a+b=1$, then the minimum value of $1 / 2a + 1 / b$ is $\qquad$
Given: $\because a, b>0$, and $a+b=1$, by Cauchy-Schwarz inequality we know:
$$\frac{1}{2 a}+\frac{1}{b}=\frac{(\sqrt{2} / 2)^{2}}{a}+\frac{1^{2}}{b} \geqslant \frac{(\sqrt{2} / 2)^{2}}{a+b}=\frac{3}{2}+\sqrt{2} \text {, }$$

with equality if and only if $\frac{\sqrt{2} / 2}{a}=\frac{1}{b}$, i.e., $a=\sqrt{2}-1, b=2-\sqrt{2}$,

thus the minimum value of $1 / 2 a+1 / b$ is $3 / 2+\sqrt{2}$.