AMC 12 Mock Exam 4 Problem 21

Write $xy+2yz=y(x+2z)$. By the Cauchy–Schwarz inequality, $|x+2z|\le\sqrt{1^2+2^2}\sqrt{x^2+z^2}=\sqrt5\sqrt{x^2+z^2}$. By AM–GM, $|y|\sqrt{x^2+z^2}\le\frac{y^2+x^2+z^2}{2}=\frac12$. Hence $xy+2yz\le\frac{\sqrt5}{2}$.

Equality needs $(x,z)$ proportional to $(1,2)$ and $y^2=x^2+z^2=\frac12$: take $x=\frac1{\sqrt{10}}$, $z=\frac2{\sqrt{10}}$, $y=\frac1{\sqrt2}$. Then $xy+2yz=\frac{1}{\sqrt2}\cdot\frac{5}{\sqrt{10}}=\frac{5}{\sqrt{20}}=\frac{\sqrt5}{2}$. The maximum is $\boxed{\tfrac{\sqrt5}{2}}$.