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Definition of the problem

I have to prove the following statement:

Let $\left(E,\left\langle \cdot,\cdot\right\rangle \right)$ be an inner product space over $\mathbb{R}$. prove that for all $x,y\in E$ we have $$ \left(\left\Vert x\right\Vert +\left\Vert y\right\Vert \right)\left|\left\langle x,y\right\rangle \right|\leq\left\Vert x+y\right\Vert \cdot\left\Vert x\right\Vert \left\Vert y\right\Vert . $$

My efforts

I tried two different ways to prove that, both unsuccessfull..

First:

First, by squaring the whole inequality:

$$ \left(\left\Vert x\right\Vert +\left\Vert y\right\Vert \right)^{2}\left|\left\langle x,y\right\rangle \right|^{2}\leq\left\Vert x+y\right\Vert ^{2}\cdot\left\Vert x\right\Vert ^{2}\left\Vert y\right\Vert ^{2}. $$ We have from Cauchy-Schwarz that $$ \left|\left\langle x,y\right\rangle \right|\leq\left\Vert x\right\Vert \cdot\left\Vert y\right\Vert $$ So we obtain $$ \left(\left\Vert x\right\Vert +\left\Vert y\right\Vert \right)^{2}\left|\left\langle x,y\right\rangle \right|^{2}\leq\left(\left\Vert x\right\Vert +\left\Vert y\right\Vert \right)^{2}\cdot\left\Vert x\right\Vert ^{2}\left\Vert y\right\Vert ^{2}=\left(\left\Vert x\right\Vert ^{2}+\left\Vert y\right\Vert ^{2}+2\left\Vert x\right\Vert \left\Vert y\right\Vert \right)\cdot\left\Vert x\right\Vert ^{2}\left\Vert y\right\Vert ^{2}. $$ By Pythagorean theorem, we obtain $$ \left(\left\Vert x\right\Vert +\left\Vert y\right\Vert \right)^{2}\left|\left\langle x,y\right\rangle \right|^{2}\leq\left(\left\Vert x+y\right\Vert ^{2}+2\left\Vert x\right\Vert \left\Vert y\right\Vert \right)\cdot\left\Vert x\right\Vert ^{2}\left\Vert y\right\Vert ^{2}. $$ We're almost there, except an extra term very annoying: $$ \left(\left\Vert x\right\Vert +\left\Vert y\right\Vert \right)^{2}\left|\left\langle x,y\right\rangle \right|^{2}\leq\left\Vert x+y\right\Vert ^{2}\cdot\left\Vert x\right\Vert ^{2}\left\Vert y\right\Vert ^{2}+2\left\Vert x\right\Vert ^{3}\left\Vert y\right\Vert ^{3}. $$

Second

I tried after to use only the Cauchy-Schwarz inequality, not squared: $$ \left(\left\Vert x\right\Vert +\left\Vert y\right\Vert \right)\left|\left\langle x,y\right\rangle \right|\leq\left(\left\Vert x\right\Vert +\left\Vert y\right\Vert \right)\cdot\left\Vert x\right\Vert \left\Vert y\right\Vert . $$

My question

Could you give me a hint/idea on how to solve this problem? which Lemma/Theorem should I use?

Thank you

Franck

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I think the statement is false. Take, for example, $E = \mathbb{R}$ with the standard inner product (just multiplication). Let $x = 1$ and $x = -1$. Then the left hand side is $2 \cdot 1 = 2$, but the right hand side is $0$. –  Rick Jun 12 '12 at 12:32
5  
Also, +1 for a well-written question with clear explanations of your attempts. –  Rick Jun 12 '12 at 12:34
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@Rick Sorry for the false problem. I will go and clarify that asap with the TA.. Thanks for your example anyway! –  FranckStudiesCommEng Jun 12 '12 at 12:42
    
However, ||x+y|| |<x,y>| <= (||x|| + ||y||) ||x|| ||y|| does hold due to Cauchy-Schwarz, and the triangle inequality (for normed spaces) –  Adam Rubinson Jun 12 '12 at 19:28
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1 Answer

up vote 4 down vote accepted

Then, let me remove this question from the dead list of "unanswered questions" by answering it.

The statement is false. A counter-example is as follows. Let $E$ be $\mathbb{R}$ itself, and the inner product be the ordinary multiplication of real numbers. Let $x = 1$ and $y = -1$. Then the left hand side is $(||x||+||y||) \cdot | \langle x, y \rangle| = (1 + 1) 1 = 2.$ The right hand side is $||x + y|| \cdot ||x|| \cdot ||y|| = ||0|| \cdot 1 \cdot 1 = 0$.

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For the ones interested, I posted a new question with the corrected statement: $$ \left(\left\Vert x\right\Vert +\left\Vert y\right\Vert \right)\left\langle x,y\right\rangle \leq\left\Vert x+y\right\Vert \cdot\left\Vert x\right\Vert \left\Vert y\right\Vert. $$ [math.stackexchange.com/questions/158901/… –  FranckStudiesCommEng Jun 16 '12 at 1:14
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