System of equations with three variables Characterize all triples $(a,b,c)$ of positive real numbers such that
$$ a^2-ab+bc = b^2-bc+ca = c^2-ca+ab. $$
This is the equality case of the so-called Vasc inequality. I think the answer is that $a=b=c$ or $a:b:c = \sin^2(4\pi/7) : \sin^2(2\pi/7) : \sin^2(\pi/7)$ and cyclic equality cases. I'm mostly interested if there is a way to derive this reasonably, other than just magically guessing the solutions and then doing degree-counting. This was left as a "good exercise" in Mildorf that has bothered me for years.

Suppose $a=b$, then $bc=b^2=c^2-cb+b^2$, those $a=b=c$. Which includes cyclic equality cases as well.
Now lets assume WLOG $a>b>c$. Show that in this case above equations don't hold simultaneously. 
