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Definition of the problem

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 $$ \left(\left\Vert x\right\Vert +\left\Vert y\right\Vert \right)\left\langle x,y\right\rangle \leq\left\Vert x+y\right\Vert \cdot\left\Vert x\right\Vert \left\Vert y\right\Vert . $$

My efforts

I tried to apply Cauchy-Schwarz inequality: $$ \left|\left\langle x,y\right\rangle \right|\leq\left\Vert x\right\Vert \left\Vert y\right\Vert \quad\forall x,y\in E, $$ and since we are in an inner product space over $\mathbb{R}$, we can simplify remove the absolute value from the inner product: $$ \left\langle x,y\right\rangle \leq\left\Vert x\right\Vert \left\Vert y\right\Vert \quad\forall x,y\in E. $$ We obtain: $$ \left(\left\Vert x\right\Vert +\left\Vert y\right\Vert \right)\left\langle x,y\right\rangle \leq\left(\left\Vert x\right\Vert +\left\Vert y\right\Vert \right)\cdot\left\Vert x\right\Vert \left\Vert y\right\Vert \quad\forall x,y\in E. $$

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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2  
Hint: remember that $||x||^2=<x,x>$ –  Gastón Burrull Jun 16 '12 at 1:14
    
Hint: what happens if x and y are colinear? If they are perpendicular? –  tomasz Jun 16 '12 at 1:18
    
If all else fails, you can always solve this kind of problem purely analytically, by assuming $x,y\in \mathbf R^2$. What i suggested pretty much boils down to this, but I withdraw that suggestion, as it is very inelegant... –  tomasz Jun 16 '12 at 1:32
1  
@tomasz I assume you would have then a loss of generality, and in consequence, would not prove the statement. –  franckysnow Jun 16 '12 at 8:34

1 Answer 1

up vote 1 down vote accepted

$\begin{align*} \big( \Vert x+y\Vert \Vert x \Vert \Vert y \Vert \big)^2 &= \Vert x + y \Vert^2 \langle x,x \rangle \langle y,y \rangle = \langle x+y,x+y\rangle \langle x,x \rangle \langle y,y \rangle = \\ &= \big(\langle x,x \rangle + \langle y,y \rangle + 2 \langle x,y \rangle \big) \langle x,x \rangle \langle y,y \rangle = \\ &= \langle x,x \rangle^2\langle y,y \rangle + \langle x,x \rangle \langle y,y \rangle^2 + 2 \langle x,y \rangle \langle x,x \rangle \langle y,y \rangle \geq \\ &\geq \langle x,y \rangle^2\langle x,x \rangle + \langle x,y \rangle^2\langle y,y \rangle + 2\langle x,y \rangle^2 \langle x,x \rangle^\frac{1}{2}\langle y,y \rangle^\frac{1}{2} = \\ &= \big( \langle x,y \rangle \langle x,x \rangle ^\frac{1}{2}+ \langle x,y \rangle \langle y,y \rangle^\frac{1}{2}\big)^2 = \big( (\langle x,x \rangle ^\frac{1}{2}+ \langle y,y \rangle^\frac{1}{2})\langle x,y \rangle \big) ^2 = \\ &= \big( (\Vert x \Vert + \Vert y \Vert)\langle x,y\rangle\big)^2 \end{align*}$

To get to the fourth line, we apply the Cauchy-Schwarz inequality to each summand.

Then, $\big( \Vert x+y\Vert \Vert x \Vert \Vert y \Vert \big)^2 \geq \big( (\Vert x \Vert + \Vert y \Vert)\langle x,y\rangle\big)^2$, and since $\Vert x+y\Vert \Vert x \Vert \Vert y \Vert \geq 0$, we can simply cancel the squares, obtaining $\Vert x+y\Vert \Vert x \Vert \Vert y \Vert \geq \big(\Vert x \Vert + \Vert y \Vert)\langle x,y\rangle$.

Note that we used $\Vert x \Vert = \langle x,x \rangle ^\frac{1}{2}$ and the fact that $\langle x+y,x+y \rangle = \langle x,x+y \rangle + \langle y,x+y\rangle =$ $=\langle x,x \rangle + \langle x,y \rangle + \langle y,x\rangle + \langle y,y \rangle = \langle x,x \rangle + \langle y,y \rangle + 2 \langle x,y \rangle$ since the inner product in a real vector space must be symmetric.

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1  
Very elegant proof, thank you! –  franckysnow Jun 16 '12 at 8:33

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