Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 3 / 23H / b / Solution

A minimal surface has $H=0$, so $k_2=-k_1$ and $\boxed{K=-k_1^2\le 0}$. For a nonflat example take the catenoid parametrization $X(u,v)=(\cosh v\cos u,\cosh v\sin u,v)$. Its first fundamental form has $E=G={\cosh }^{2}v$ and $F=0$. Choosing the normal appropriately, its second fundamental form has coefficients $e=-1$, $f=0$, $g=1$. Thus its principal curvatures are $\pm {\mathrm{sech}}^{2}v$, proving $H=0$ and $\boxed{K=-{\mathrm{sech}}^{4}v<0}$.