 Solution

The Cauchy-Schwarz inequality in ${\mathbb{R}}^n$ isHere the Euclidean norm is induced by the dot product. If $b=0$, the result is immediate. If $b\ne 0$, the squared Euclidean norm of the component of $a$ perpendicular to $b$ is nonnegative:Rearrangement proves the Cauchy-Schwarz inequality. Equality holds precisely when $a$ and $b$ are linearly dependent, including the cases of a zero vector.

Expanding the squared Euclidean norm and using the Cauchy-Schwarz inequality givesTaking nonnegative square roots gives the triangle inequality. Applying the triangle inequality once more gives