Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 2 / 24H / Solution

The plane planar isoperimetric inequality says that a simple closed curve of length $L$ enclosing area $A$ satisfies $L^2\ge 4\pi A$, with equality for circles.

For sufficiently small $r$, below the injectivity radius at $p$, geodesic polar coordinates give the surface metric $d{r}^2+j(r,\theta {)}^{2}d{\theta }^2$. The radial variation field is a Jacobi field, soIts 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, giveThe Gauss lemma makes the angular metric factor both the circle line element and the polar area factor. ThereforeSquaring the length expansion and subtracting givesThis identifies the requested remainder and proves its ratio to $r^4$ tends to zero. For $K(p)>0$, sufficiently small geodesic disks have $L^2<4\pi A$, 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.