$a,\,b,\,c\ge 0$
Ta có: $\begin{cases}(\sqrt{a}-\sqrt{b})^2\ge 0\\(\sqrt{b}-\sqrt{c})^2\ge 0\\(\sqrt{a}-\sqrt{c})^2\ge 0\end{cases}$
$⇔(\sqrt{a}-\sqrt{b})^2+(\sqrt{b}-\sqrt{c})^2+(\sqrt{a}-\sqrt{c})^2\ge 0$
$⇔a-2\sqrt{ab}+b+b-2\sqrt{bc}+c+a-2\sqrt{ac}+c\ge 0$
$⇔2(a+b+c)-2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})\ge 0$
$⇔2(a+b+c)\ge 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})$
$⇔a+b+c\ge \sqrt{ab}+\sqrt{bc}+\sqrt{ac}$ (Đpcm).