Difference between revisions of "Cauchy-Schwarz inequality"

Latest revision as of 13:04, 4 April 2017

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
• [math]\vert\langle x,y\rangle\vert\le\Vert x\Vert \Vert y\Vert[/math] - 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 [math]a_1,...,a_n,b_1,...,b_n\in\mathbb{R}\ [/math] we will have
[math]\sum^n_{i=1}a_ib_i\le\sqrt{\sum^n_{i=1}a_i^2}\sqrt{\sum^n_{i=1}b_i^2}[/math]

Proof

Basis for argument

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

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

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

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

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

Conclusion of first argument

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

Core of argument

In the basis we required a function, [math]f(x)[/math], we will now build this.

Take [math]\sum^n_{i=1}(a_it+b_i)^2[/math] and notice:

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

Using the above this means [math]b^2-4ac\le 0[/math], where:

• [math]a=\sum^n_{i=1}a_i^2[/math]
• [math]b=2\sum^n_{i=1}a_ib_i[/math]
• [math]c=\sum^n_{i=1}b_i^2[/math]

Conclusion of argument

[math]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[/math][math]\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)[/math][math]\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}[/math]

But as [math]x\le|x|[/math] (recall [math]|\cdot|[/math] denotes absolute value) we see:

[math]\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}[/math]

QED