<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] We need to prove that $$c^2+ab\leq(a+b)c$$ or
$$a^2+b^2\leq(a+b)\sqrt{a^2-ab+b^2}$$ or
$$a^2+b^2\leq\sqrt{(a+b)(a^3+b^3)},$$ which is true by C-S.
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