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?
 A: 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.
A: 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.
A: Form condition we get $c = \sqrt{a^2-ab+b^2},$ therefore
$$(a-c)(b-c)=a^2+b^2-c(a+b)=a^2+b^2-(a+b)\sqrt{a^2-ab+b^2}$$
$$=-\frac{ab(a-b)^2}{a^2+b^2+(a+b)\sqrt{a^2-ab+b^2}} \leqslant 0.$$
A: By the law of sines, we have that
$$a^{2}+ b^{2}- 2ab\cdot\sin 60^{\circ}= c^{2}\Rightarrow c:={\rm med}\left \{ a, b, c \right \}\Rightarrow \left ( a- c \right )\left ( b- c \right )\leq 0$$
A: 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',
$(a-c)(b-c)\leq0$
