<s>[INST] Let $x$, $y$, $z$ be three positive reals such that $x+y+z=\sqrt{10+\sqrt{19}}$ and $\frac1x+\frac1y+\frac1z=\sqrt{10-\sqrt{19}}$ ... 
Let $x,y,$ and $z$ be three positive real numbers such that
$x+y+z=\sqrt{10+\sqrt{19}}$ and
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\sqrt{10-\sqrt{19}}$. If
$a=x^2+y^2+z^2$ and $b=a+\frac{9}{a}$, find $b$.

Solution:
Just take the product and you get $\sum_{cyc}\frac{x}{y}+\frac{y}{x}=6$ after simplifying. Note that by AM-GM on LHS we have $\sum_{cyc}\frac{x}{y}+\frac{y}{x}\ge 3(2)=6$, and because equality occurs in AM-GM iff all the variables are equal we find that $\frac{x}{y}=\frac{y}{x}\Leftrightarrow x=y$, and cyclically. Hence $x=y=z$ so from the first equation we get $x=\frac{\sqrt{10+\sqrt{19}}}{3}$. Plugging into $a$ gives $a=3x^2=\frac{10+\sqrt{19}}{3}$, and plugging into $b$ yields $b=a+\frac{9}{a}=\frac{10+\sqrt{19}}{3}+\frac{27}{10+\sqrt{19}}=\frac{20}{3}$.
Question: Why is $x = y = z $ in this solution?
 [/INST] Let's write $\sum_{cyc} \frac{x}{y}+\frac{y}{x}$ explicitly as $\frac{x}{y}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{z}{x}+\frac{x}{z}$ and denote this by $A$. Then the arithmetic mean of these $6$ numbers is $\frac{A}{6}$ and the geometric mean of these $6$ numbers is $\sqrt[6]{1}=1$. By AM-GM we have $\frac{A}{6}\geq 1$ with equality if and only if all $6$ numbers are equal, that is, $A\geq 6$ with equality if and only if $\frac{x}{y}=\frac{y}{x}=\frac{y}{z}=\ldots$. Since it was shown before that indeed $A=6$ then all $6$ numbers must be equal and in particular $\frac{x}{y}=\frac{y}{x}$ therefore $x=y$, and similarly for $y=z$.
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