Cauchy-Schwarz inequality

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Statement

For any a1,...,an,b1,...,bnR  we will have
ni=1aibini=1a2ini=1b2i

Proof

Basis for argument

Consider first the function f:RR give by f(x)=ax2+bx+c

If f(x)0 then using the quadratic equation we know the solutions (to f(x)=0) will at be: x=b±b24ac2a

As we want f(x)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 b24ac=0 as then b±b24ac2a=b±02a=b2a - our 2 repeated solutions.

In the second case we require b24ac<0 as then the b24ac term will be imaginary, thus giving us no real solutions.

Conclusion of first argument

We conclude from this that if a quadratic ax2+bx+c is to be 0 then b24ac0

Core of argument

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

Take ni=1(ait+bi)2 and notice:

  1. ni=1(ait+bi)2=ni=1(a2it2+2taibi+b2i)=t2ni=1a2i+2tni=1aibi+ni=1b2i - which is a quadratic in t
  2. ai,bi,tR (ait+bi)20, so ni=1(ait+bi)20 - our quadratic in t is 0

Using the above this means b24ac0, where:

  • a=ni=1a2i
  • b=2ni=1aibi
  • c=ni=1b2i

Conclusion of argument

4(ni=1aibi)24(ni=1a2i)(ni=1b2i)0(ni=1aibi)2(ni=1a2i)(ni=1b2i)|ni=1aibi|ni=1a2ini=1b2i

But as x|x| (recall || denotes absolute value) we see:

ni=1aibini=1a2ini=1b2i

QED

[<collapsible-expand>]Real Analysis