A problem on Minima. Proving sphere has minimum surface area for a given volume

I recommend reading the first part of The Brunn-Minkowski inequality for nilpotent groups by Terence Tao. He first considers the simple one-dimensional Brunn-Minkowski inequality, then proves the Prékopa-Leindler inequality, which again implies the full version of Brunn-Minkowski:

Brunn-Minkowski inequality. Let $A,B$ be two nonempty bounded open subsets of $\mathbb{R}^d$. Denote the Minkowski sum $$ A+B = \{ a+b : a \in A, \ b \in B \} $$ and let $|\cdot|$ be the $n$-dimensional Lebesgue measure. Then $$ |A+B|^n \ge |A|^{1/n} + |B|^{1/n}. $$

The isoperimetric inequality follows. If $A \subseteq \mathbb{R}^d$ has the same volume $\omega_n$ as the unit ball and $B = B(0,r)$, then $A+B$ equals $A_r$ - the $r$-neighborhood of $A$. Thus Brunn-Minkowski reads $$ |A_r|^{1/n} \ge |\omega_n|^{1/n} + |\omega_n r^n|^{1/n} = \omega_n^{1/n} (1+r), $$ then by taking the $n$-th power and substracting $|A|$, $$ \frac{|A_r|-|A|}{r} \ge \frac{\omega_n ((1+r)^n - 1)}{r}. $$ If we now take $r \to 0$, the left-hand side converges to the area of $\partial A$ if only $A$ is sufficiently regular (or if we take this as the definition of area), while the right-hand side converges to $n \omega_n$, which is the area of the unit ball. Hence $$ \mathrm{Area}(\partial A) \ge \mathrm{Area}(\partial B(0,1)). $$

Of course this doesn't show that the sphere is the unique minimizer. The discussion linked by G. Sassatelli deals with this issue.