60.2. If $a$ and $b$ are positive real numbers, what is the minimum value of the expression
$$
\sqrt{a+b}\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}\right) ?
$$
60.2. $2 \sqrt{2}$
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
$$
\sqrt{a+b}\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}\right) \geq 2 \sqrt{2} .
$$