Answer
See solution
Work Step by Step
$|\langle\vec{u}, \vec{v}\rangle|=|a+b|$
$||\vec{u}||=\sqrt{ \langle\vec{u}, \vec{u}\rangle }=\sqrt{a^2+b^2}$
$||\vec{v}||=\sqrt{ \langle\vec{v}, \vec{v}\rangle }=\sqrt{2}$
By the Cauchy-Schwarz Inequality, $|a+b|\le \sqrt{2(a^2+b^2)}$
Squaring both sides, $(a+b)^2\le 2(a^2+b^2)$
Dividing both sides by 4, we get the desired inequality
$\left(\frac{a+b}{2}\right)^2\le \frac{a^2+b^2}{2}$
