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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?

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  • $\begingroup$ i think this is an inequality with $a,b,c>0$ $\endgroup$ – Dr. Sonnhard Graubner Jan 1 '18 at 20:35
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The hint.

Let $b=a+u$,$c=a+v$ and $u=xv$.

Hence, $$(a^2+b^2+c^2)^2-3(a^3b+b^3c+c^3a)=\sum_{cyc}(a^4+2a^2b^2-3a^3b)=$$ $$=(u^2-uv+v^2)a^2+(u^3-5u^2v+4uv^2+v^3)a+u^4-3u^3v+2u^2v^2+v^4\geq0$$ because $$(u^3-5u^2v+4uv^2+v^3)^2-4(u^2-uv+v^2)(u^4-3u^3v+2u^2v^2+v^4)=$$ $$=-3(u^3-u^2v-2u^2+v^3)^2=-3v^6(x^3-x^2-2x+1)^2\leq0.$$ The equality occurs for $x^3-x^2-2x+1=0$, which gives $$x\in\left\{2\cos\frac{\pi}{7},-2\cos\frac{2\pi}{7},2\cos\frac{3\pi}{7}\right\}.$$ For example, $x=2\cos\frac{\pi}{7}$ gives the following case. $$\frac{b}{\sin^2\frac{4\pi}7}=\frac{c}{\sin^2\frac{2\pi}7}=\frac{a}{\sin^2\frac{\pi}7}.$$ My solution by $uvw$ see here: https://artofproblemsolving.com/community/c6h6026p5329091

There is also the following nice solution. https://gbas2010.wordpress.com/2010/01/08/problem-19vasile-cirtoaje/

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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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  • $\begingroup$ But $a$, $b$ and $c$ are reals and why they are all solutions? $\endgroup$ – Michael Rozenberg Jan 1 '18 at 20:42
  • $\begingroup$ See Dr. Sonnhard Graubner's comment. $\endgroup$ – Mohammad Zuhair Khan Jan 2 '18 at 7:49

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