Problem - 2071
Let $A=x\cos^2{\theta} + y\sin^2{\theta}$, $B=x\sin^2{\theta}+y\sin^2{\theta}$, where $x$, $y$, $A$, and $B$ are all real numbers. Prove $x^2 + y^2 \ge A^2 + B^2$
Let $z_1=A+Bi$, $z_2=x+yi$. Then
$|z_1|=|(x\cos^2{\theta} + y\sin^2{\theta})+(x\sin^2{\theta}+y\sin^2{\theta})i|=|z_2\cos^2{\theta}+z_2\sin^2{\theta}|\le |z_2\cos^2{\theta}|+|z_2\sin^2{\theta}| = |z_2|$
The means $|z_2|^2 \ge |z_1|^2$, i.e. $A^2 + B^2 \ge x^2 + y^2$.