<s>[INST] Prove That $x=y=z$ If $x, y,z \in \mathbb{R}$,
and if
$$ \left ( \frac{x}{y} \right )^2+\left ( \frac{y}{z} \right )^2+\left ( \frac{z}{x} \right )^2=\left ( \frac{x}{y} \right )+\left ( \frac{y}{z} \right )+\left ( \frac{z}{x} \right ) $$
Prove that $$x=y=z$$
 [/INST] Let $p = a+b+c$, then $ab+bc+ca = \frac{1}{2}(p^2 - p)$ and $abc=1$. So $a,b,c$ are solutions to the equation
$$x^3 - px^2 + \frac{1}{2} (p^2-p)x - 1 = 0$$
The discriminant is $(\frac{1}{2}(p^2-p))^2 - 4(\frac{1}{2}(p^2-p))^3 - 4p^3 - 27 + 18 p (\frac{1}{2}(p^2-p)) \ge 0$, since the roots are real.
This is equivalent to 
$$p^6 - 4p^5 + 5p^4 - 22p^3 + 36p^2 + 108 \leq 0$$
But by calculus or other means, one can check that the left hand side has minimum 0 only at $p = 3$. Thus $p = 3$, and $ab+bc+ca = 3$, $abc = 1$. So $a,b,c$ are roots of $x^3-3x^2+3x-1 = (x-1)^3$, i.e. $a,b,c = 1$, i.e. $x=y=z$.
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