$a,b,c$ are positive reals and distinct with $a^2+b^2 -ab=c^2$. Prove $(a-c)(b-c)<0$ 
$a,b,c$ 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.
 A: If $a<b$, then
$$
c^2 = a^2+b^2-ab = a^2+b(b-a) > a^2 \\
c^2 = a^2+b^2-ab = b^2-a(b-a) < b^2
$$
which means $a<c<b$ and $(a-c)(b-c)<0$.
If $a>b$, then
$$
c^2 = a^2+b^2-ab = a^2-b(a-b) < a^2 \\
c^2 = a^2+b^2-ab = b^2+a(a-b) > b^2
$$
which means $b<c<a$ and $(a-c)(b-c)<0$.
A: Observe triangle with sides $a,b,c$ and angles $\alpha,\beta, \gamma$. 
By cosine theorem angle $\gamma = 60^{\circ}$, thus $\alpha +\beta =120^{\circ}$. 
So we can assume $\alpha \leq 60^{\circ}$ and $\beta \geq 60^{\circ}$. So $b\geq c\geq a$ and thus conclusion:
$$(a-c)(b-c)\leq 0$$
with equality iff $a=b=c$.
A: WLOG assume $a<b$. Then:
$$a^2+b^2-ab=c^2 \Rightarrow a(a-b)=(c-b)(c+b) \Rightarrow c<b$$
$$a^2+b^2-ab=c^2 \Rightarrow (a-c)(a+c)=b(a-b) \Rightarrow a<c.$$
A: Let $a\le b$
$a,b,c$ - sides of $\triangle ABC$ with $\angle BCA= 60^o$ , $BC=a , AC=b$.
$\Rightarrow a\le c\le b \Rightarrow (a-c)(b-c)\le 0$
A: we get $$c=\sqrt{a^2+b^2-ab}$$ since $c>0$ and $$a^2+b^2>ab$$ from here we get
the product
$$(a-\sqrt{a^2+b^2-ab})(b-\sqrt{a^2+b^2-ab})<0$$
we have two cases.
1) $$a>\sqrt{a^2+b^2-ab}$$ and $$b<\sqrt{a^2+b^2-ab}$$ after squaring we get $$a>b$$
2) $$a<\sqrt{a^2+b^2-ab}$$ and $$b>\sqrt{a^2+b^2-ab}$$
after squaring we get $$a<b$$ and our Statement is true.
A: Without loss of generality you can supose $a>b$. Then there are 2 cases you got to disprove (sorry if this word does not exist, english is not my first language)
Case 1) $a>b>c$ and Case 2) $c>a>b$.
Any other case, satisfies the proppert (Since a,b, and c are dsitinct)
Case 1)
If $a>b>c$ then $b^2>c^2$ and $a^2-ab > 0 $ (since a > b).
Then $a^2 +b^2 -ab > c^2 + 0$. Then it contradicts the hypothesis.
Case 2)
If $c>a>b$ then $c^2>a^2$ and $0>b^2-ab$ then $c^2=c^2+0>a^2+b^2-ab. Contradicting the hypothesis again.
Keep in mind this can all be done only because they ae positive real numbers.
