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 $$x=y=z$$
 A: 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. $x=y=z$.
A: Let $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$, and let
$$
G=\sqrt[3]{abc}, A = \frac{a+b+c}{3}, Q=\sqrt{\frac{a^2+b^2+c^2}{3}}
$$
Then the given condition is $Q^2=A$, but by the power mean inequality
$$
Q^2\ge Q \ge A \ge G = 1
$$
with equality in each case only if $a=b=c=1$, i.e. if $x=y=z$.
A: Let $$ a=\frac{x}{y} $$ $$ b=\frac{y}{z} $$ $$ c=\frac{z}{x} $$
Then $$abc=1$$
$$ a^2+b^2+c^2=a+b+c$$
which implies
$$ (a-0.5)^2+(b-0.5)^2+(c-0.5)^2=0.75$$
I was struck up here, i would appreciate if any one helps me here
A: 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 $f(x,y)\ge f(1,1)=3$.
