# Inner product space law

• September 14th 2008, 12:51 PM
Inner product space law
In an inner product space, show that $||x+y|| ||x-y|| \leq ||x||^2 + ||y||^2$

Proof.

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

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

But I don't seem to be able to get to it, am I starting this wrong? Thanks.
• September 14th 2008, 02:01 PM
Opalg
Quote:

Originally Posted by tttcomrader
In an inner product space, show that $||x+y|| ||x-y|| \leq ||x||^2 + ||y||^2$

Try squaring both sides, then expand $\|x\pm y\|^2$ as $\langle x\pm y,x\pm y\rangle$.