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.

Note that $2ab > -ab \Rightarrow a^2+b^2+2ab > a^2 + b^2 -ab \Rightarrow (a+b)^2 > a^2 + b^2 -ab \Rightarrow a+b > \sqrt{a^2 + b^2 -ab}$
I was able to do this by squaring the original inequality I was supposed to prove, and then I worked backwards. 
