Symmetrization and isoperimetric inequality Let $\Omega\subseteq\mathbb{R}^n$ be a bounded open set with $C^1$ boundary $\partial\Omega$. The isoperimetric inequality states $$\frac{|\partial\Omega|}{|\Omega|^{\frac{n-1}{n}}}\geq \frac{|\partial B_1|}{|B_1|^{\frac{n-1}{n}}},$$ where $B_1$ is the unit ball centred at $0$.
I want to prove a weaker version, say that given a fixed volume $V$ the ball has the lowest perimetre among all the regions $\Omega$ with volume $|\Omega|=V$. (*)
My professor told me to focus on an $\Omega$ of the form
$$ \Omega=\{(x',x_n)\in\mathbb{R}^{n-1}\times\mathbb{R}:\,x'\in W\subseteq\mathbb{R}^{n-1},\,\varphi_1(x')<x_n<\varphi_2(x')\}, $$ and then to consider $$\Omega_s=\{(x',x_n)\in\mathbb{R}^{n-1}\times\mathbb{R}:\,x'\in W\subseteq\mathbb{R}^{n-1},\,-\frac{\varphi_2(x')-\varphi_1(x')}{2}<x_n<\frac{\varphi_2(x')-\varphi_1(x')}{2}\}.$$
Direct computations show that $|\Omega|=|\Omega_s|$ and $|\partial\Omega|\geq |\partial\Omega_s|$, with equality if and only if $\Omega$ is symmetric with respect to a hyperplane $\{x_n=C\}$. 
I understand that this provides an intuition for (*). 
However, I do not see how to extend this idea in order to achieve a formal proof of (*).
 A: So far, you have shown that the operation $\Omega \mapsto \Omega_s$, which provides a set with the same volume as $\Omega$, but symmetric to a given hyperplane decreases the perimeter. 
In order to profe the isoperimetric inequality in this form, you need to do the following:
(This proof mimics the strategy used when dealing with arbitrary measurable sets (of finite perimeter). There are possibly other approaches more suitable for dealing with sets with $C^1$ boundary.) 


*

*Show that a perimeter minimizing set (for given volume) actually exists. Call that minimizer $M$ Here you need some sort of compactness property of the space of considered sets. Currently I cannot provide you an easy argument. 

*As you already proved (for a special case), symmetrizing a set with respect to a given hyperplane decreases the perimeter. Thus $|\partial M_s|\le |\partial M| $ and as $|\partial M|$ is already minimial $|\partial M_s|= |\partial M| $.

*We now need to show, that the above implies $M=M_s$, i.e. $M$ is symmetric with respect to the chosen hyperplane.
I think if you only consider sets with $C^1$ boundary, this follows from the computation.
In the general case, you need to first show that the above implies that $M$ is convex.   If $M$ is convex, you can parametrize it in the same form as your $\Omega$, with $\varphi_1$ convex, $\varphi_2$ concave (thus they are differentialbe almost everywhere). An explicit calculation of the perimeter then shows that $\varphi_1= -\varphi_2+c$ for some constant $c\in \mathbb R$. But this means that up to translation $M$ is equal to $M_s$

*Since the hyperplane with repect to which we symmetrized in the beginning was arbitrary, it follows that $M$ is symmetric with respect to all hyperplanes. But then $M$ is a ball.

