# Thread: Inner product space law

1. ## Inner product space law

In an inner product space, show that $\displaystyle ||x+y|| ||x-y|| \leq ||x||^2 + ||y||^2$

Proof.

I want to show that $\displaystyle 2||x||^2 + 2||y||^2 \geq 2||x+y||||x-y||$

By the Parallelogram law, I know that $\displaystyle 2||x||^2 +2||u||^2 = ||x+y||^2+||x-y||^2 \geq ||x+y||^2-||x-y||^2$$\displaystyle =( ||x+y|| + ||x-y||)(||x+y|| - ||x-y||)$

But I don't seem to be able to get to it, am I starting this wrong? Thanks.

In an inner product space, show that $\displaystyle ||x+y|| ||x-y|| \leq ||x||^2 + ||y||^2$
Try squaring both sides, then expand $\displaystyle \|x\pm y\|^2$ as $\displaystyle \langle x\pm y,x\pm y\rangle$.