Suppose $x,y,z$ are positive real numbers that satisfy \begin{align*} \frac{x}{y} + \frac{y}{x} + \frac{x}{z} + \frac{z}{x} + \frac{y}{z} + \frac{z}{y} &= 2018 \\ \sqrt{\frac{x}{y}} + \sqrt{\frac{y}{z}} + \sqrt{\frac{z}{x}} &= 17. \end{align*} Find the value of $\sqrt{\frac{y}{x}} + \sqrt{\frac{z}{y}} + \sqrt{\frac{x}{z}}.$
I wasn't sure how to start on this problem as there weren't any good factorizations that I could use.
Squaring the second equation we get
$\frac{x}{y} + \frac{y}{z} + \frac{z}{x} + 2(\sqrt{\frac{x}{z}} + \sqrt{\frac{z}{y}} + \sqrt{\frac{y}{x}}) = 289.$
Subtracting from the first equation $\frac{x}{z} + \frac{z}{y} + \frac{y}{x} - 2(\sqrt{\frac{x}{z}} + \sqrt{\frac{z}{y}} + \sqrt{\frac{y}{x}}) = 2018-289 = 1729.$
Put $a = \sqrt{\frac{y}{x}} + \sqrt{\frac{z}{y}} + \sqrt{\frac{x}{z}}.$
$$a^2 = \frac{x}{z} + \frac{z}{y} + \frac{y}{x} + 2(\sqrt{\frac{x}{y}} + \sqrt{\frac{y}{z}} + \sqrt{\frac{z}{x}}) = 1729+2a+34 = 1763+2a$$
This is a quadratic equation in a and has the positive solution a=43.
