I was looking for a simple proof by induction of the isoperimetric inequality, the work of Emmanuel Tsukerman gives a promising approach.
There is however one line where I cannot follow the reasoning, here is an excerpt of the proof (page 4 in the link above)
Theorem 1. (Classical Isoperimetric Inequality) For any body $K$ with n-dimensional volume $|K|$ and surface area $|\partial K|$, $$\frac{|K|^{n-1}}{|\partial K|^n}\le\frac{|B^n|^{n-1}}{|\partial B^n|^n}$$ Proof. The base case of $n=1$ is straightforward, since the boundary $\partial K$ of a closed interval has volume $|\partial K|=2$ (the counting measure for two boundary points of a line segment) and so does the boundary of the ball. Assume by induction that the inequality holds for dimension $n-1$. Let $K\subset \mathbb R^n$ and $\partial K$ its boundary.
Without loss of generality, we may assume that $|K|=|B^n|$. Define $K_t=K\cap \{x_n=t\}$ and $\partial K_t=\partial K\cap \{x_n=t\}$. Finally, let $V(t)=|K_t|$ and $A(t)=|\partial K_t|$. Note that because each $K_t$ is a parallel slice, $\int V(t)\mathrm dt =|K|$. Then $$V'(t)=\int_{\partial K_t}\frac{1}{\tan\theta}$$ where $\theta$ denotes the angle formed by the $x_n$-axis and the unit normal vector to $\partial K$.
What is the reasoning behind the following line?
$$V'(t) = \int_{ \partial K_t} \frac{1}{\tan \theta} $$ where $\theta$ denotes the angle formed by the $x_n$-axis and the unit normal vector to $\partial K$.