<s>[INST] 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?
 [/INST] WLOG, we assume that $a \le b$ and we want to show that
$$a \le \sqrt{a^2+b^2-ab}\le b$$
which is equivalent to
$$a^2\le a^2+b^2-ab \le b^2$$
The first inequality reduces to $0 \le b(b-a)$ which is clearly true.
The second inequality is $a^2-ab=a(a-b) \le 0$ which is true again.
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