Prove that $||x-y||·||x+y|| < ||x||^2+||y||^2$

Question

I have a doubt to prove this exercise:

$$||x-y||·||x+y|| < ||x||^2+||y||^2 \;\;\;\;\; \forall\; x,y \in R^n$$

What I've done:

$$||x-y||·||x+y|| = \sqrt{(x-y,x-y)(x+y,x+y)} = \sqrt{((x,x)+(-y,-y))((x,x)+(y,y)}) = \sqrt{((x,x)+(y,y))^2} = (x,x)+(y,y) = ||x||^2+||y||^2$$

But then, $||x-y||·||x+y|| = ||x||^2+||y||^2$, not $<$.

Why?

Thanks in advance

Answer 1

Let us do it by parts. To start with, we have

\begin{align*} \lVert x-y\rVert^{2} = \langle x - y,x - y\rangle = \lVert x\rVert^{2} - 2\langle x,y\rangle + \lVert y\rVert^{2} \end{align*} Analogously, \begin{align*} \lVert x+y\rVert^{2}= \langle x + y,x + y\rangle = \lVert x\rVert^{2} + 2\langle x,y\rangle + \lVert y\rVert^{2} \end{align*} which implies that \begin{align*} \lVert x-y\rVert^{2}\times\lVert x+y\rVert^{2} = (\lVert x\rVert^{2} + \lVert y\rVert^{2})^{2} - 4\langle x,y\rangle^{2} \leq (\lVert x\rVert^{2} + \lVert y\rVert^{2})^{2} \end{align*} from whence the result holds.