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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 $$x=y=z$$

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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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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

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The triple completing-the-square is a great technique, but it doesn't look like the way to go here, partly because $0.75$ is disappointing, but mostly because you didn't use $abc=1$. – vadim123 Apr 25 '13 at 3:54

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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Apply AMGM directly, you get $x+y+\frac{1}{xy} \ge 3$. – user27126 Apr 25 '13 at 4:16
D'oh! Thanks @Sanchez – vadim123 Apr 25 '13 at 4:16

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 $x=y=z$.

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