Let $\mu(z) dV\_n$ be a measure in $\mathbb{C} ^n$.
Let $B\_n(r) := \{z \mid \|z\| < r\}$ be the ball of radius $r$ in $\mathbb C^n$, and $\partial B\_n(r) $ be the corresponding sphere.
In $\mathbb{C} $ how can we find the following inequality?
$$
\operatorname{Vol}\_{\mu}(B\_1(r))=\int\_{B\_1(r)} \mu(z) dV\_1(z)=
\int\_0^r\left(\int\_{\partial B\_1(t)} \mu dz\right)dt\geq \int\_0^r \left[\int\_{\partial B\_1(t)}(\mu)^{ \frac{1}{2}} \right]^2\frac{1}{2\pi t} dt
$$
And can we generalize this inequality in $\mathbb {C}^n$?



Relationship between volume and area
The problem can be solved via co-area formula and Jensen's inequality. We will do it Bourbaki style, i.e from $n$-dimensional case to particular case $n=1$.




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Instead of $\mathbb C^n$, we can equivalently see the problem as a problem in $\mathbb R^m$, where $m=2n$ (i.e we isomorphically map real dimensions for each complex dimension). So, let $dV\_m$ denote volume measure in $\mathbb R^m$ and $dS\_{m-1}$ denote the corresponding surface area measure, i.e $(m-1)$-dimensional Hausdorff measure. The mapping $F: z \mapsto \|z\|$ on $\mathbb R^m$ has jacobian determinant $1$ (except at $z=0$, where it is undefined). Also note that for all $t \ge 0$, we have
$$
F^{-1}(\{t\})=\{z \in \mathbb R^m \mid F(z) = t\} = \{z \in \mathbb R^m \mid \|z\| = t\} = \partial B\_m(t).
$$
By the coarea-formula (see [Corollary 1.4](https://www3.nd.edu/%7Elnicolae/Coarea.pdf), for example), we have
$$
\begin{split}
\int\_{B\_m(r)}\mu(z)dV\_m(z) &= \int\_{0}^r\left(\int\_{F^{-1}(\{t\})}\frac{\mu(z)}{|Jac\_F(z)|}dS\_{m-1}(z)\right)dt\\
&= \int\_{0}^r\left(\int\_{\partial B\_m(t)}\mu(z)dS\_{n-1}(z)\right)dt\\
&= \int\_{0}^r\left(\int\_{\partial B\_m(t)}\frac{\mu(z)}{S\_{m-1}(\partial B\_m(t))}dS\_{m-1}(z)\right)S\_{n-1}(\partial B\_m(t))dt\\
&\ge\int\_{0}^r\left(\int\_{\partial B\_m(t)}\frac{\mu(z)^{1/2}}{S\_{m-1}(\partial B\_m(t))}dS\_{m-1}(z)\right)^2S\_{m-1}(\partial B\_m(t))dt\\
&= \int\_{0}^r\left(\int\_{\partial B\_m(t)}\mu(z)^{1/2}dS\_{m-1}(z)\right)^2\frac{1}{S\_{m-1}(\partial B\_m(t))}dt,
\end{split}
$$
where the inequality is an applicaiton of Jensen's inequality on the convex function $x \mapsto x^2$ and the probability measure $A \mapsto S\_{m-1}(A \cap \partial B(t))/S\_{m-1}(\partial B(t))$.


In particular, if $n=2$, we have $m=2\cdot 1 = 2$, $dS\_{m-1} = dS\_1$ which is the arc-length measure, and so $S\_1(\partial B(t)) =$ length or circle of radius $t$, which equals $2\pi t$.


