Clarifying the proof of Cauchy-schwarz inequality

Question

Why on earth is the 2.term >= the 3.term and why the absolute values are suddenly inside the sum notation

Thanks to everyone who helps

Answer 1

Let me write out Cauchy-Schwarz for two terms.

$$|A_1B_1+A_2B_2| \le \sqrt{A_1^2+A_2^2}\sqrt{B_1^2+B_2^2}$$

Let me rewrite it:

$$ \sqrt{A_1^2+A_2^2}\sqrt{B_1^2+B_2^2}\ge |A_1B_1+A_2B_2|$$

Now, let me write $A_1=\sqrt{\sum_{i=1}^ka_i^2}, A_2=|a_{k+1}|,B_1=\sqrt{\sum_{i=1}^kb_i^2}, B_2=|b_{k+1}| $

We have

\begin{align} &\sqrt{\sum_{i=1}^k a_i^2 + a_{k+1}^2}\sqrt{\sum_{i=1}^k b_i^2 + b_{k+1}^2} \\&=\sqrt{A_1^2+A_2^2}\sqrt{B_1^2+B_2^2}\\&\ge|A_1B_1 +A_2B_2| \end{align}

by Cauchy-Schwarz for $2$-terms.

The second inequality is due to Cauchy-Schwarz for $k$ terms. Try to verify this.