# Cauchy-Schwarz inequality

There are two forms of this inequality:

• [ilmath]\sum^n_{i=1}a_ib_i\le\sqrt{\sum^n_{i=1}a_i^2}\sqrt{\sum^n_{i=1}b_i^2}[/ilmath] - the common and
• $\vert\langle x,y\rangle\vert\le\Vert x\Vert \Vert y\Vert$ - the rare but more general

TODO: More general version http://math.stackexchange.com/questions/1357968/cauchy-schwarz-inequality-proof-but-not-the-usual-one

Update: Cauchy-Schwarz inequality for inner product spaces is a proof of the second form - note that [ilmath]\Vert x\Vert:\eq\sqrt{\langle x,x\rangle} [/ilmath] is the norm induced by the inner product Alec (talk) 13:04, 4 April 2017 (UTC)

## Statement

For any $a_1,...,a_n,b_1,...,b_n\in\mathbb{R}\$ we will have
$\sum^n_{i=1}a_ib_i\le\sqrt{\sum^n_{i=1}a_i^2}\sqrt{\sum^n_{i=1}b_i^2}$

## Proof

### Basis for argument

Consider first the function $f:\mathbb{R}\rightarrow\mathbb{R}$ give by $f(x)=ax^2+bx+c$

If $f(x)\ge 0$ then using the quadratic equation we know the solutions (to $f(x)=0$) will at be: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

As we want $f(x)\ge 0$ we must have either a repeated solution (a point where $f(x)=0$) or no real solutions.

In the first case (repeated solutions) we require $b^2-4ac=0$ as then $\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-b\pm0}{2a}=\frac{-b}{2a}$ - our 2 repeated solutions.

In the second case we require $b^2-4ac<0$ as then the $\sqrt{b^2-4ac}$ term will be imaginary, thus giving us no real solutions.

#### Conclusion of first argument

We conclude from this that if a quadratic $ax^2+bx+c$ is to be $\ge0$ then $b^2-4ac\le 0$

### Core of argument

In the basis we required a function, $f(x)$, we will now build this.

Take $\sum^n_{i=1}(a_it+b_i)^2$ and notice:

1. $\sum^n_{i=1}(a_it+b_i)^2=\sum^n_{i=1}(a_i^2t^2+2ta_ib_i+b_i^2)=t^2\sum^n_{i=1}a_i^2+2t\sum^n_{i=1}a_ib_i+\sum^n_{i=1}b_i^2$ - which is a quadratic in $t$
2. $\forall a_i,b_i,t\in\mathbb{R}\ (a_it+b_i)^2\ge 0$, so $\sum^n_{i=1}(a_it+b_i)^2\ge0$ - our quadratic in $t$ is $\ge0$

Using the above this means $b^2-4ac\le 0$, where:

• $a=\sum^n_{i=1}a_i^2$
• $b=2\sum^n_{i=1}a_ib_i$
• $c=\sum^n_{i=1}b_i^2$

### Conclusion of argument

$4\left(\sum^n_{i=1}a_ib_i\right)^2-4\left(\sum^n_{i=1}a_i^2\right)\left(\sum^n_{i=1}b_i^2\right)\le 0$$\iff\left(\sum^n_{i=1}a_ib_i\right)^2\le\left(\sum^n_{i=1}a_i^2\right)\left(\sum^n_{i=1}b_i^2\right)$$\iff\left|\sum^n_{i=1}a_ib_i\right|\le\sqrt{\sum^n_{i=1}a_i^2}\sqrt{\sum^n_{i=1}b_i^2}$

But as $x\le|x|$ (recall $|\cdot|$ denotes absolute value) we see:

$\iff\sum^n_{i=1}a_ib_i\le\sqrt{\sum^n_{i=1}a_i^2}\sqrt{\sum^n_{i=1}b_i^2}$

QED