Prove inequality using AM-GM inequality. $$\sqrt{a}+\sqrt{b}\geq \sqrt{a+b}$$ I know it is trivial to prove straight forward but I need to prove it using AM-GM inequality.

\begin{align*}
\sqrt a + \sqrt b
&= \frac{(\sqrt a + \sqrt b + \sqrt2 \sqrt[4]{ab}) +
          (\sqrt a + \sqrt b - \sqrt2 \sqrt[4]{ab})}{2} \\
&\ge \sqrt{\big(\sqrt a + \sqrt b + \sqrt2 \sqrt[4]{ab}\big)
          \big(\sqrt a + \sqrt b - \sqrt2 \sqrt[4]{ab}\big)} \\
&= \sqrt{\big(\sqrt a + \sqrt b\big)^2 - 2 \sqrt{ab}} \\
&= \sqrt{a+b}
\end{align*}
(To apply AM/GM thus requires $\sqrt a + \sqrt b \ge \sqrt2\sqrt[4]{ab}$; happily, $\sqrt a + \sqrt b \ge 2\sqrt[4]{ab}$, by AM/GM again.)
