= Solution
The plane <planar isoperimetric inequality> says that a simple closed curve of length $L$ enclosing area $A$ satisfies $L^2\geq4\pi A$, with equality for circles.
For sufficiently small $r$, below the <injectivity radius> at $p$, <geodesic polar coordinates> give the surface metric $dr^2+j(r,\theta)^2d\theta^2$. The radial variation field is a <Jacobi field>, so
$$
j_{rr}+K(\gamma_\theta(r))j=0,\qquad j(0,\theta)=0,\quad j_r(0,\theta)=1.
$$
Its integral equation is $j=r-\int_0^r(r-u)K(\gamma_\theta(u))j(u,\theta)\,du$. <Continuity> of <curvature> and $j(u,\theta)=u+o(u)$, uniformly in direction, give
$$
j(r,\theta)=r-\frac{K(p)}6r^3+o(r^3).
$$
The <Gauss lemma> makes the angular metric factor both the circle line element and the polar area factor. Therefore
$$
L=\int_0^{2\pi}j(r,\theta)\,d\theta
=2\pi r-\frac\pi3K(p)r^3+o(r^3),
$$
$$
A=\int_0^r\int_0^{2\pi}j(u,\theta)\,d\theta\,du
=\pi r^2-\frac\pi{12}K(p)r^4+o(r^4).
$$
Squaring the length expansion and subtracting gives
$$
\boxed{4\pi A-L^2=\pi^2K(p)r^4+o(r^4).}
$$
This identifies the requested remainder and proves its ratio to $r^4$ tends to zero. For $K(p)>0$, sufficiently small <geodesic> disks have \b[$L^2<4\pi A$], the opposite strict inequality to the planar bound. Their positive <curvature> allows more area for the same perimeter; the Euclidean inequality is not a universal inequality for curved surfaces.
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