TITLE: How to prove this inequality in Banach space?
QUESTION [8 upvotes]: In a normed space $(E,\lVert \cdot\rVert)$ space we have the following inequality:
$$\forall\, x,y\in E,\quad\|x\|^{2}-\|y\|^{2}\leq \lVert x-y\rVert\cdot \|x+y\|.$$
How can we prove it?

REPLY [14 votes]: We have
$$2||x||=||2x||=||x+y+x-y||\leq||x+y||+||x-y|| $$
so 
$$4||x||^2\leq ||x+y||^2+2||x+y||\cdot||x-y||+||x-y||^2,$$
and 
$$|||x+y||-||x-y|||\leq 2||y||$$
so 
$$||x+y||^2-2||x+y||\cdot||x-y||+||x-y||^2\leq 4||y||^2.$$
We get 
\begin{align*}4(||x||^2-||y||^2)&\leq ||x+y||^2+2||x+y||\cdot||x-y||+||x-y||^2\\
&-(||x+y||^2-2||x+y||\cdot||x-y||+||x-y||^2)\\
&=4||x+y||\cdot||x-y||,
\end{align*}
hence $||x||^2-||y||^2\leq ||x+y||\cdot||x-y||$ for all $x,y$.
(we don't need a Banach space, a normed space is enough)