# Cauchy-Schwarz Inequality Proof

• Feb 18th 2011, 05:04 PM
ragnar
Cauchy-Schwarz Inequality Proof
So I'm reading this document that has a proof of the Cauchy-Schwarz Inequality, and I'm missing the logic.

It says, "By Proposition 2.3 and (i) of Proposition 2.5, we have (assuming $y \ne 0$, otherwise, nothing needs to be proved)" (The propositions referenced basically establish the linearity of dot-product, and that the norm is always non-negative.)

$0 \leq ||x - \lambda y||^{2} = ||x||^{2} - 2 \lambda x \cdot y + \lambda^{2} ||y||^{2} \leq (||x|| - \lambda ||y||)^{2} + 2\lambda\big( ||x|| \, ||y|| - |x\cdot y|\big)$.

I get everything here except the last inequality. I see that it reduces to this:

$-2 \lambda x \cdot y \leq -2 \lambda |x \cdot y|$ but I don't see why I'm supposed to believe this.

Also, tangentially, can anyone tell me what's wrong with this proof?:
$||x||\, ||y|| = (x \cdot x)^{1/2} (y \cdot y)^{1/2} = (x \cdot x \cdot y \cdot y)^{1/2} = (x \cdot y)^{1/2}(x \cdot y)^{1/2} = |x \cdot y|$
• Feb 18th 2011, 05:56 PM
Ackbeet
Hmm. You have

$\|x\|^{2} - 2 \lambda x \cdot y + \lambda^{2} \|y\|^{2} \le (\|x\| - \lambda \|y\|)^{2} + 2\lambda\big( \|x\| \, \|y\| - |x\cdot y|\big).$

The RHS is

$(\|x\| - \lambda \|y\|)^{2} + 2\lambda\big( \|x\| \, \|y\| - |x\cdot y|\big)=(\|x\|-\lambda\|y\|)(\|x\|-\lambda\|y\|)+2\lambda\big( \|x\| \, \|y\| - |x\cdot y|\big)$

$=\|x\|^{2}-2\lambda\|x\|\,\|y\|+\lambda^{2}\|y\|^{2}+2\lambda \|x\|\,\|y\|-2\lambda|x\cdot y|=\|x\|^{2}-2\lambda|x\cdot y|+\lambda^{2}\|y\|^{2}.$

So you are correct in that it does boil down to whether

$-2\lambda\,x\cdot y\le-2\lambda|x\cdot y|.$

That inequality is most definitely false. Counterexample:

$x=\langle 1,0\rangle,\quad y=\langle -1,0\rangle.$

Then let's suppose $\lambda>0.$ We'd have

$2\lambda=-2\lambda(-1+0)=-2\lambda x\cdot y\overset{?}{\le}-2\lambda|x\cdot y|=-2\lambda|-1+0|=-2\lambda.$

For positive $\lambda,$ this does not hold. Therefore, I'd say this inequality of the proof is invalid, and you were right to doubt it.

See here for a valid proof.

Your proof, unfortunately, doesn't work because the expression

$x\cdot x\cdot y\cdot y$

doesn't parse mathematically. The result of a dot product is a scalar. The dot product of two scalars is not defined (unless you mean ordinary multiplication in the real numbers).

You should also be suspicious of your proof, because you've proven too much. Is it really true that

$\|x\|\,\|y\|=|x\cdot y|?$

Plug in two orthonormal vectors, and you can see that the equality won't work. The LHS is 1, and the RHS is zero.
• Feb 19th 2011, 01:51 AM
tonio
Quote:

Originally Posted by ragnar
So I'm reading this document that has a proof of the Cauchy-Schwarz Inequality, and I'm missing the logic.

It says, "By Proposition 2.3 and (i) of Proposition 2.5, we have (assuming $y \ne 0$, otherwise, nothing needs to be proved)" (The propositions referenced basically establish the linearity of dot-product, and that the norm is always non-negative.)

$0 \leq ||x - \lambda y||^{2} = ||x||^{2} - 2 \lambda x \cdot y + \lambda^{2} ||y||^{2} \leq (||x|| - \lambda ||y||)^{2} + 2\lambda\big( ||x|| \, ||y|| - |x\cdot y|\big)$.

I get everything here except the last inequality. I see that it reduces to this:

$-2 \lambda x \cdot y \leq -2 \lambda |x \cdot y|$ but I don't see why I'm supposed to believe this.

Also, tangentially, can anyone tell me what's wrong with this proof?:
$||x||\, ||y|| = (x \cdot x)^{1/2} (y \cdot y)^{1/2} = (x \cdot x \cdot y \cdot y)^{1/2} = (x \cdot y)^{1/2}(x \cdot y)^{1/2} = |x \cdot y|$

After your first equality, and assuming $\lambda\in\mathbb{R}$ , we get $||x||^{2} - 2 \lambda x \cdot y + \lambda^{2} ||y||^{2} \geq 0$

This is a quadratic in $\lambda$ which is always non-negative, so its discriminant must be non-positive:

$\Delta=4(x\cdot y)^2-4||x||^2||y||^2\leq 0\Longleftrightarrow |x\cdot y|\leq ||x||||y||$ , and this is the C.S. inquality.

I know of no proof of the C.S. ineq. easiest than the above.

Tonio
• Feb 19th 2011, 07:47 AM
PaulRS
Here's a proof by normalization (if we are working over $\mathbb{R}$ ) :

The inequality is trivial if $x = \bold{0}$ or $y = \bold{0}$, so assume they are non-zero.

We have: $\langle v-w,v-w\rangle \geq 0$ and $\langle v+w,v+w\rangle \geq 0$

Now that means that no matter what the sign of $\langle v, w \rangle$ is, we have that: $\langle v, v\rangle + \langle w, w\rangle \geq 2 \cdot \left | \langle v, w\rangle \right |$

Thus, let $v = \frac{x}{\|x\|}$ and $w = \frac{y}{\|y\|}$ to get : $2 = \frac{\langle x, x \rangle}{\|x\|^2} + \frac{\langle y, y \rangle}{\|y\|^2} \geq 2 \cdot \left | \left\langle \frac{x}{\|x\|}, \frac{y}{\|y\|} \right\rangle \right |$
• Feb 19th 2011, 07:58 AM
Quote:

Originally Posted by ragnar
So I'm reading this document that has a proof of the Cauchy-Schwarz Inequality, and I'm missing the logic.

It says, "By Proposition 2.3 and (i) of Proposition 2.5, we have (assuming $y \ne 0$, otherwise, nothing needs to be proved)" (The propositions referenced basically establish the linearity of dot-product, and that the norm is always non-negative.)

$0 \leq ||x - \lambda y||^{2} = ||x||^{2} - 2 \lambda x \cdot y + \lambda^{2} ||y||^{2} \leq (||x|| - \lambda ||y||)^{2} + 2\lambda\big( ||x|| \, ||y|| - |x\cdot y|\big)$.

I get everything here except the last inequality. I see that it reduces to this:

$-2 \lambda x \cdot y \leq -2 \lambda |x \cdot y|$ but I don't see why I'm supposed to believe this.

Also, tangentially, can anyone tell me what's wrong with this proof?:
$||x||\, ||y|| = (x \cdot x)^{1/2} (y \cdot y)^{1/2} = (x \cdot x \cdot y \cdot y)^{1/2} = (x \cdot y)^{1/2}(x \cdot y)^{1/2} = |x \cdot y|$

By the way , what document is that ? isn't it the "Cauchy-Schwartz master class" ?
I will be very grateful if you share it because i use this kind of inequalities a lot .
Thanks .