Prove That $x=y=z$

Solution 1:

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. $x=y=z$.

Solution 2:

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 $x=y=z$.

Solution 3:

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