# Gauss curvature of graphs of a function

Suppose that $f,g:\mathbb{R}^{2}\rightarrow \mathbb{R}$ are smooth functions, with $g(u,v)\geq f(u,v)\geq 0$ for all $u, v \in\mathbb{R}$ and with f(0,0)=g(0,0)=0. Let $κ_{f},κ_{g}$ be respectively the Gauss curvatures of the graphs of f and g at the origin. Show that $κ_{g}≥κ_{f}≥0$.

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Since $g$ and $f$ agree up to first-order derivatives, the inequality $g\ge f$ implies the corresponding inequality for Hessians: $D^2g \ge D^2f$ in the sense of positive semidefinite order. The inequality then propagates to eigenvalues (ordered, of course): $\lambda_j(D^2g)\ge \lambda_j(D^2f)$ for $j=1,\dots,n$. The proof for $n=2$ is not hard: see the comment by Jonas Meyer here.