(A local Isoparametric inequality for geodesic circles) Let $p\in S$ and let $S_{r}(p)$ be a geodesic circle of center $p$ and radius $r$ . Let $L$ be the arc length of $S_{r}(p)$ and $A$ be the area of the region bounded by $S_{r}(p)$ . Prove that $$4\pi -L^2=\pi^2r^4K(p)+R,$$
where $K(p)$ is the Gaussian curvature of $S$ at $p$ and $$\lim_{r\to 0}\frac{R}{r^4}=0$$ Thus, if $K(p)>0$ $(\text{ or } <0)$ and $r$ is small, $4\pi A-L^2 >0$$(\text{ or } <0)$.
I assume you know \begin{equation} \frac{1}{2\pi}L(r)=r-\frac{K}{3!}r^3+o(r^3)\tag{1}\label{eqn:L} \end{equation} as $r\to 0$, is one way to define the curvature $K(p)=K$ at a point $p$ on a surface.
Integrating \eqref{eqn:L}, \begin{equation} \frac{1}{2\pi}A=\frac12r^2-\frac{K}{4!}r^4+o(r^4)\tag{2}\label{eqn:A} \end{equation} because we can interpret $L(r)\,\mathrm{d}r$ as the infinitesimal increase of area from radius $r$ to $r+\,\mathrm{d}r$.
Substitute \eqref{eqn:L} and \eqref{eqn:A} into $4\pi A-L^2$, the coefficient of $r^2$ vanish as expected, and the next term comes out to be $\pi^2Kr^4$. The error term is obvious.