Prove $\|x-y\|\|x+y\|\le\|x\|^2+\|y\|^2$ for all $x,y\in\mathbb{R}^n$ Prove $\|x-y\|\|x+y\|\le\|x\|^2+\|y\|^2$  for all $x,y\in\mathbb{R}^n$
I've been struggling with this for a while and haven't figured out a way to do it either geometrically or algebraically. 

\begin{align}
\|x-y\|^2\|x+y\|^2 &=(x-y,x-y)(x+y,x+y)\\
&=(\|x\|^2-2(x, y) + \|y\|^2)(\|x\|^2+2(x,y) + \|y\|^2)\\
&=(\|x\|^2+\|y\|^2)^2-4(x, y)^2\\
&\le(\|x\|^2+\|y\|^2)^2
\end{align}
