Solution
= Solution
A <minimal surface> has $H=0$, so $k_2=-k_1$ and $\boxed{K=-k_1^2\leq0}$. 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 $E=G=\cosh^2v$ and $F=0$. Choosing the normal appropriately, its <second fundamental form> has coefficients $e=-1$, $f=0$, $g=1$. Thus its principal curvatures are $\pm\operatorname{sech}^2v$, proving $H=0$ and $\boxed{K=-\operatorname{sech}^4v<0}$.