Tenth Philippine Mathematical Olympiad

Solution:
$2 \sqrt{2}$

By the AM-GM Inequality, we have
$$\sqrt{a+b} \geq \sqrt{2 \sqrt{a b}} = \sqrt{2} (a b)^{1 / 4}$$
and
$$\frac{1}{\sqrt{a}} + \frac{1}{\sqrt{b}} \geq 2 \sqrt{\frac{1}{\sqrt{a}} \cdot \frac{1}{\sqrt{b}}} = \frac{2}{(a b)^{1 / 4}}$$
where both inequalities become equalities if and only if $a = b$. Multiplying the two inequalities, we get
$$\sqrt{a+b}\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}\right) \geq 2 \sqrt{2}$$