Comparing Surface Areas

$\sqrt{r^2+h^2}<\sqrt{r^2+h^2+(3h^2+4rh)}=(r+2h)$ $(\because 3h^2+4rh>0)$ $\implies \sqrt{r^2+h^2}<r+2h$ $\implies \pi r\sqrt{r^2+h^2}<\pi r(r+2h)$ $\implies \pi r^2+\pi r\sqrt{r^2+h^2}<\pi r^2+\pi r(r+2h)$ $\implies \pi r^2+\pi r\sqrt{r^2+h^2}<2\pi r^2+2\pi rh$ $\implies$ Surface area of cone $<$ Surface area of cylinder.

Note : In other words, we want to find the mathematical symbol satisfying the inequation/equation below.

$2 \pi r^2 + 2\pi r h \stackrel{?}{=} \pi r^2 + \pi rs \; ,$

where $s$ denote the slant length of the cone. By Pythagorean theorem , $s$ satisfy the equation, $r^2 + h^2 = s^2$ , so $s = \sqrt{r^2+ h^2}$ .