= Solution
Put $\mathbf z=(x,y)^T$, $r^2=\mathbf z^T\mathbf z$ and $f=\sqrt{1+r^2}$. The <chain rule> yields
$$
\boxed{\nabla f=\frac{\mathbf z}{f},\qquad H_f=\frac I f-\frac{\mathbf z\mathbf z^T}{f^3}=\frac1{f^3}\begin{pmatrix}1+y^2&-xy\\-xy&1+x^2\end{pmatrix}.}
$$
For every nonzero <vector> $\mathbf v$, the <Cauchy-Schwarz inequality> gives
$$
\mathbf v^TH_f\mathbf v=\frac{(1+r^2)\|\mathbf v\|^2-(\mathbf z^T\mathbf v)^2}{f^3}\geq\frac{\|\mathbf v\|^2}{f^3}>0.
$$
Thus the <Hessian> is a <positive-definite matrix> everywhere. The <Hessian criterion for convexity> implies that \b[$f$ is strictly <convex> on $\mathbb R^2$]. Its unique global <minimum> is at $\mathbf z=0$, where $f=1$.
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