Prove Cauchy-Schwarz inequality by induction

This page proves the Cauchy-Schwarz inequality

\[ \sqrt{\sum_{i=1}^n a_i^2} \sqrt{\sum_{j=1}^n b_i^2} \ge \sum_{i=1}^n a_i b_i \]

by induction.

Proof

Case \(n=2\)

If there are only two variables in each sum,

\begin{align} (a_1^2+a_2^2)(b_1^2+b_2^2)-(a_1b_1+a_2b_2)^2&= a_1^2b_1^2+ a_1^2b_2^2+ a_2^2b_1^2+ a_2^2b_2^2- (a_1^2b_1^2+2a_1b_1a_2b_2+a_2^2b_2^2) \\ &=a_1^2b_2^2+a_2^2b_1^2-2a_1b_1a_2b_2 \\ &=(a_1b_2-a_2b_1)^2 \\ &\ge0. \end{align}

Case \(n+1\)

Suppose the theorem is true for \(n\), so

\[ (a_1^2+a_2^2+...+a_n^2)(b_1^2+b_2^2+...+b_n^2)\ge (a_1b_1+a_2b_2+...+a_nb_n)^2 \]

then let \(\alpha=\sqrt{a_1^2+a_2^2+...+a_n^2}\) and \(\beta=\sqrt{b_1^2 +...+ b_n^2}\). These are well defined because the sums of squares are always greater than or equal to zero.

By supposition, \begin{align} a_1b_1+a_2b_2+...+a_nb_n+a_{n+1}b_{n+1} &\le \alpha\beta+a_{n+1} b_{n+1} \\ &\le \sqrt{\alpha^2 + a_{n+1}^2}\sqrt{\beta^2 + b_{n+1}^2} \\ &= \sqrt{a_1^2 +...+ a_{n+1}^2}\sqrt{b_1^2 +...+ b_{n+1}^2} \end{align}

where the case \(n=2\) is used in the second inequality.

Acknowledgement

This proof is based on this post at math.stackexchange.