This question is related to Proof by induction of n-dimensional isoperimetric inequality, missing step. Discussing the same inequality but with a different approach. Gary Lawlor shows in his 2010 print An elementary isoperimetric proof in $n$-space a proof by induction of
Theorem 2.1. Round circles, spheres and hyperspheres are the most efficient volume in all dimensions; for any surface in $\mathbb{R}^n$ enclosing volume $V$, the surface area $S$ satisfies $$ S \geq \frac{nV}{r},$$ where $r$ is the radius of the round sphere enclosing the same volume $V$.
Below is a verbatim copy of the author's proof up until the point where I cannot follow anymore.
Proof: Let $T$ be the surface of some solid $B \subset \mathbb{R}^n$. [...] We will use terminology familiar to $\mathbb{R}^3$, namely "volume" for $n$-volume, "area" for ($n-1)$-volume, and "perimeter" or "length" for $(n-2)$-volume. [...] Define $r$ so that the volume $V$ of $B$ matches that of a round $n$-ball of radius $r$. [...] For each $t$ let $A(t)$ be the area of the planar slice $B \cap \{x_n =t\}$. Let $P(t)$ be its perimeter, the length of $T \cap \{x_n = t\}.$ Let $t_1$ and $t_2$ be the lowest and highest values of $x_n$ on $T$, and let $S$ be the surface area of $T$.
Under the above setup the author mentions the basic inequality for surface area: \begin{align} S \geq \int_{t_1}^{t_2} \sqrt{P(t)^2 + A'(t)^2 }dx_n. \tag{*} \end{align}
My question: How does one prove (*)? If possible, I'd like to see an approach via geometric measure theory, in particular with the help of the coarea or area formula. Although I haven't been exposed to this theory myself I find it the easiest to self-study.
My approach: For the sake of completeness I give my own attempt to establish (*). Using:
Theorem (Volume of Submanifolds) Let $n \leq m$ and $f: \mathbb{R}^n \to \mathbb{R}^m$ be a Lipschitz injective map, $A \subset \mathbb{R}^n$ Lebesgue measurable (of finite measure), then $$ \mathcal{H}^n(f(A))= \int_A Jf(y)dy. $$
with $n=1, m=2$ and $A= [t_1,t_2] \subset \mathbb{R}$. We define $f: A \to \mathbb{R}^2$ as \begin{align*} f(t):= \begin{pmatrix} \int_{t_1}^t P(s) ds \\ A(t) \end{pmatrix}. \end{align*} Then $f$ can be seen to be Lipschitz injective (using that the solid is bounded, sufficiently smooth boundary) and $Jf(t)= \sqrt{ P(t)^2 + A'(t)^2}$, in particular we have recreated the RHS of (*). However, the identity $$ \mathcal{H}^1(f([t_1,t_2]))= \int_{t_1}^{t_2} \sqrt{P(t)^2 + A'(t)^2}dt $$ doesn't seem to be very helpful, since $\mathcal{H}^1$ is just the "length" of the curve $f([t_1,t_2])$.