If $a,b,c$ are positive integers, with $a^2+b^2-ab=c^2$ prove that $(a-b)(b-c)\le0$. I have an inequality problem which is as follow:
If $a,b,c$ are positive integers, with $a^2+b^2-ab=c^2$ prove that $(a-b)(b-c)\le0$.
I am not so good in inequalities. So, please give me some hints so that I can proceed.
Thanks.

The expression $ a^2 + b^2 - ab = c^2 $ is equivalent to
$$
a^2 + b^2 - ab = c^2 \longleftrightarrow a^2 - ab = c^2 - b^2 \longleftrightarrow a(a - b) = (c - b)(c + b) 
$$
Now we multiply by $ b - c $ both sides (if $ b - c = 0 $ then there is nothing to prove):
$$
a(a - b)(b - c) = (c - b)(c + b)(b - c) = -(c - b)^2 (c + b)
$$
The right side is negative, because $ -(c - b)^2 \leq 0 \ $ (a square is always positive) and  $ c,b \geq 0 \rightarrow c + b \geq 0 $.
Then we deduce 
$$ 
a(a - b)(b - c) \leq 0 
$$
And we know that $ a \geq 0 $, so
$$
(a - b)(b - c) \leq 0 
$$
