View source for Cauchy-Schwarz inequality

There are two forms of this inequality:
* {{M|1=\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
* {{MM|1=\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 }}

:: {{Highlight|Update: [[Cauchy-Schwarz inequality for inner product spaces]] is a proof of the second form - note that {{M|\Vert x\Vert:\eq\sqrt{\langle x,x\rangle} }} is the [[norm induced by the inner product]] [[User:Alec|Alec]] ([[User talk: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<br/>
<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|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|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:
# <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>
# <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|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


[[Category:Useful inequalities]]
{{Theorem Of|Real Analysis}}