If $a^2+b^2-ab=c^2$ for positive $a$, $b$, $c$, then show that $(a-c)(b-c)\leq0$ 
Let $a$, $b$, $c$ be positive numbers. If $a^2+b^2-ab=c^2$. Show that
$$(a-c)(b-c)\leq0$$

I have managed to get the equation to $(a-b)^2=c^2-ab$, but I haven't been able to make any progress.
Can someone help me?

Form condition we get $c = \sqrt{a^2-ab+b^2},$ therefore
$$(a-c)(b-c)=a^2+b^2-c(a+b)=a^2+b^2-(a+b)\sqrt{a^2-ab+b^2}$$
$$=-\frac{ab(a-b)^2}{a^2+b^2+(a+b)\sqrt{a^2-ab+b^2}} \leqslant 0.$$
