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?

Equation, $(a^2-ab+b^2)=(c)^2$ has solution shown below:
$a=p^2-q^2$
$b=2pq-q^2$
$c=p^2-pq+q^2$
Where, $p>q>0$
We need:
$w=(a-c)(b-c)\leq0$
Or, $w=(c-a)(c-b)\leq0$
$(c-b)=(p-q)(p-2q)$
$(c-a)=-q(p-2q)$
Hence,
$w=(c-a)(c-b)=(-)(p-q)(q)(p-2q)^2$
Since, $p>q>0$:
$(p-2q)^2$ & $(p-q)$ & $(q)$ is a positive quantity and the expression 'w' is negative,
The expression 'w',
$(a-c)(b-c)\leq0$
