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.

From given equality we have $c=\sqrt {a^2+b^2-ab}$, then you can plug this in your inequality and you get $(a-b)(b-\sqrt{a^2+b^2-ab})$, now you have 2 cases to look at:
First: $a>b, a=b+x$ with this your first term is positive, so we have to show second one is negative, thus we have $b-\sqrt{b^2+bx+x^2};\sqrt{b^2+bx+x^2} > \sqrt {b^2}$, it is obvious from here that the product is negative.
Second: $a<b,a=b-x$, now we know that first term is negative, so we have to show that second one is positive, again we plug in and get $b-\sqrt{b^2-bx-x^2};\sqrt{b^2-bx-x^2} < \sqrt {b^2}$, so we know that this one is positive, so the product is again negative.
And finally if $a=b$ then their product is $0$ because $a-b=0$.
