Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Discrete mathematics, continuous mathematics, and probability resit / 6 / b / Solution

Put $\mathbf{z}=(x,y{)}^T$, $r^2={\mathbf{z}}^T\mathbf{z}$ and $f=\sqrt{1+r^2}$. The chain rule yieldsFor every nonzero vector $\mathbf{v}$, the Cauchy-Schwarz inequality givesThus the Hessian is a positive-definite matrix everywhere. The Hessian criterion for convexity implies that $f$ is strictly convex on ${\mathbb{R}}^2$. Its unique global minimum is at $\mathbf{z}=0$, where $f=1$.