<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] A similar triple completing-the-square is helpful.  Using your notation, $(a-1)^2+(b-1)^2+(c-1)^2+2(a+b+c)-3=a+b+c$.  This rearranges to $(a-1)^2+(b-1)^2+(c-1)^2=3-(a+b+c)$.  Hence $a+b+c\le 3$ since otherwise the sum of three squares would be negative.  Consequently $a^2+b^2+c^2\le 3$.
Unfortunately, I don't have a cute trick for the opposite inequality.  Let $x=a^2, y=b^2$.  I want to minimize $f(x,y)=x+y+\frac{1}{xy}$, assuming $x,y>0$.  Setting the partials equal to zero, I get $1-\frac{1}{xy^2}=0=1-\frac{1}{yx^2}$, which has unique solution $x=y=1$.  (as $x,y$ approach either 0 or $\infty$, $f(x,y)$ grows without bound, so this is a minimum).  Consequently $f(x,y)\ge f(1,1)=3$.
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