Cauchy-Schwarz inequality

There are two forms of this inequality:

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

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

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Update: Cauchy-Schwarz inequality for inner product spaces is a proof of the second form - note that $\Vert x\Vert:\eq\sqrt{\langle x,x\rangle}$ 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