Why is Fourier analysis so handy for proving the isoperimetric inequality?

Experience with Fourier analysis and representation theory has shown that every time a problem is invariant with respect to a group symmetry, the representation theory of that group is likely to be relevant. If the group is abelian, the representation theory is given by the Fourier transform on that group.

In this case, the relevant symmetry group is that of reparameterising the arclength parameterisation of the perimeter by translation. This operation does not change the area or the perimeter. When combined with the observation (from Stokes theorem) that both the area and perimeter of a body can be easily recovered from the arclength parameterisation, this naturally suggests to use Fourier analysis in the arclength variable.

It seems to me that it is worth mentioning that only the 2-d isoperimetric inequality is easily proved using Fourier analysis but not higher dimensional geometric inequalities. In addition to the group invariance property cited by Terry Tao, there is the simple fact that a closed curve can be represented by a pair of periodic functions, and, if the curve is parameterized properly, its length and area are nicely represented by integrals of quadratic polynomials of the periodic functions and their derivatives. All in all, a nice setup for Fourier series. If the integrands were higher degree polynomials, a proof might still be possible but I'm not sure it would be as easy. And is there an analogous proof in higher dimensions?

This doesn't answer the question of why Fourier analysis works, but it certainly is an answer to how one might think "Hmm ... perhaps we're in the domain of Fourier analysis here." It's that the surface area of a shape X is defined in terms of the volume of X+B when B is a small ball. There is a close relationship between sumsets and convolutions (the sumset is precisely the set of points where the convolution of the characteristic functions of the two sets is non-zero), and every time you have a convolution, thoughts of Fourier analysis should be triggered.

The reason I say this doesn't answer the question of why Fourier analysis works is that there is a difference between the convolution and the support of the convolution, and the latter does not transform nicely. But that shouldn't stop one thinking of Fourier analysis and attempting to find some way of relating surface area to convolutions.