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?

By the law of sines, we have that
$$a^{2}+ b^{2}- 2ab\cdot\sin 60^{\circ}= c^{2}\Rightarrow c:={\rm med}\left \{ a, b, c \right \}\Rightarrow \left ( a- c \right )\left ( b- c \right )\leq 0$$
