TITLE: $a,b,c$ are positive reals and distinct with $a^2+b^2 -ab=c^2$. Prove $(a-c)(b-c)<0$
QUESTION [5 upvotes]: $a,b,c$ are positive reals and distinct with $a^2+b^2 -ab=c^2$.
Show that $(a-c)(b-c)<0$.

This is a question presented in the "Olimpiadas do Ceará 1987" a math contest held in Brazil. Sorry if this a duplicate.
Given the assumptions, it is easy to show that
$$0<(a-b)^2<a^2+b^2-ab=c^2.$$ But could not find a promising route to pursue.
Any hint or answer is welcomed.

REPLY [2 votes]: Let $a\le b$
$a,b,c$ - sides of $\triangle ABC$ with $\angle BCA= 60^o$ , $BC=a , AC=b$.
$\Rightarrow a\le c\le b \Rightarrow (a-c)(b-c)\le 0$