<s>[INST] How to solve (in)equality of three variables with trigonometric solutions I'm working through a set of inequality problems and I'm stuck on the following question: 

Find all sets of solutions for which $$(a^2+b^2+c^2)^2=3(a^3b+b^3c+c^3a)$$ holds. Note that $a,b,c\in\mathbb{R}$.

Firstly, one can easily see that when $a=b=c$ then the equality holds: $$\text{LHS}=(a^2+a^2+a^2)^2=(3a^2)^2=9a^4$$ and $$\text{RHS}=3(a^4+a^4+a^4)=3(3a^4)=9a^4.$$
A hint is given to use the substitutions $a=x+2ty$, $b=y+2tz$ and $c=x+2tz$ for real $t$. The LHS is rather nice in that it simplifies to $$(4t(xy+xz+yz)+(1+4t^2)(x^2+y^2+z^2))^2$$ but I can't find a similar simplification for the RHS. I guess this is a type of uvw question but I don't know where to start.
There are actually four sets of solutions: $$a=b=c,$$ $$\frac{a}{\sin^2\frac{4\pi}7}=\frac{b}{\sin^2\frac{2\pi}7}=\frac{c}{\sin^2\frac{\pi}7},$$ $$\frac{b}{\sin^2\frac{4\pi}7}=\frac{c}{\sin^2\frac{2\pi}7}=\frac{a}{\sin^2\frac{\pi}7},$$ and $$\frac{c}{\sin^2\frac{4\pi}7}=\frac{a}{\sin^2\frac{2\pi}7}=\frac{b}{\sin^2\frac{\pi}7}$$ I have no idea how the  trigonometric expressions are obtained. How could the equality be solved?
 [/INST] Hint:
Take the square root of all terms in $\frac{c}{\sin^2\frac{4\pi}7}=\frac{a}{\sin^2\frac{2\pi}7}=\frac{b}{\sin^2\frac{\pi}7}$.
Now you obtain a case of a well known equation: the sine rule.
$\frac {4\pi}7 + \frac {2\pi}7 +\frac {\pi}7= \pi$, so this is a triangle with side lengths $\sqrt a, \sqrt b$ and $\sqrt c$ and angles $\frac {\pi}7, \frac {2\pi}7 $ and $ \frac {4\pi}7$.  
Hope you can take it from here!
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