Chapter 6 - Orthogonality and Least Squares - 6.7 Exercises - Page 385: 19

See solution

$|\langle\vec{u}, \vec{v}\rangle|=2\sqrt{ab}$ $||\vec{u}||=\sqrt{ \langle\vec{u}, \vec{u}\rangle }=\sqrt{a+b}$ $||\vec{v}||=\sqrt{ \langle\vec{v}, \vec{v}\rangle }=\sqrt{b+a}$ By the Cauchy-Schwarz Inequality, $2\sqrt{ab}\le\sqrt{a+b}\sqrt{b+a}$, which means $\sqrt{ab}$ is smaller than or equal to $\frac{a+b}{2}$