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.

$$a+b> \sqrt{a^2+b^2-ab} \iff (a+b)^2>\left (\sqrt{a^2+b^2-ab}\right )^2$$  $$ \iff\quad a^2+2ab+b^2>a^2+b^2-ab$$ $$\iff\quad 3ab>0 \quad \iff\quad ab>0$$ which is correct because of the hypothesis $a,b >0$.
