Cauchy-Schwarz inequality

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Statement

For any a1,...,an,b1,...,bnR  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