Prove a variant of the Cauchy-Schwarz inequality

If $\bar a$ and $\bar b$ are vectors in $\mathbb{R^n}$, prove that $| \bar a \cdot \bar b | \leq || \bar a|| || \bar b||$ given that $|| \bar b||=1$. Well, I intuitively know that if $||\bar b||=1$, then $\bar b$ is a unit vector because the norm is 1. Since in this case $| \bar a \cdot \bar b | \leq || \bar a||$, the question boils down to basically proving $a_1b_1+a_2b_2+...\leq \sqrt{a_1^2+a_1^2+...}$. What should I do or manipulate to solve this problem?

Solution 1:

We want to use the inequality,

For $a, b\ge 0$

$0\le(a-b) ^2$

$\implies ab\le \frac{a^2}{2} + \frac{b^2}{2}$

Let, $a=(a_1, a_2,..., a_n) $ $b=(b_1, b_2,..., b_n) $

Assume $a\neq 0$ otherwise it would be trivial.

Choose, $x_i=\frac{|a_i|}{||a||}$

And $y_i=|b_i|$

Then, $x_i, y_i \ge 0$ , applying the above inequality,

$|x_i||y_i|\le \frac{|x_i|^2}{2} + \frac{|y_i^2|}{2}$

$\implies (\frac{|a_i|}{||a||})(|b_i|) \le \frac{|a_i|^2}{2 ||a||^2} +\frac{|b_i^2|}{2}$

Now, \begin{align} \sum_{i=1}^{n}(\frac{|a_i|}{||a||})(|b_i|)&\le \sum_{i=1}^{n}\frac{|a_i|^2}{2||a||}+\sum_{i=1}^{n}\frac{(|b_i|)^2}{2} \\ &=\frac{||a||^2}{2 ||a||^2} +\frac {||b||^2}{2}\\ &=\frac{1}{2}+\frac{1}{2}\\ &=1 \end{align}

Hence,$ |a•b|\le ||a||$