How to show that $a+b> \sqrt{a^2+b^2-ab}, \qquad a, b >0$ How do you show that 
$$a+b> \sqrt{a^2+b^2-ab}, \qquad a, b >0$$
I could write $\sqrt{a^2+b^2-ab}=\sqrt{(a+b)^2-3ab}$, but this seems to lead nowhere.

$3ab>0$ 
Hence, $2ab>-ab$
Hence, $a^2+2ab+b^2>a^2+b^2-ab$
Hence $a+b>\sqrt{a^2+b^2-ab}$ because $a^2+b^2-ab>a^2+b^2-2ab=(a-b)^2>0$
