Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question

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)

share|cite|improve this answer
@Julian Thanks for editing! – Davide Giraudo Dec 18 '11 at 17:12

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.