Taken at face value, the question as it stands contains incorrect / inconsistent / confusing data (see my comment). Below, I will try to answer my interpretation of the problem...

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**Preamble.** I'll do the analysis in $\mathbb R^m$, for $m \in \{1,2,3,\ldots\}$, hoping that things will work similarly in any $\mathbb C^n$, since $\mathbb C^n$ is more or less the same thing as $\mathbb R^{2n}$.

So, let $dV_m$ denote volume measure in $\mathbb R^m$ and $dS_{m-1}$ denote the surface area measure. The mapping $F: z \mapsto \|z\|$ on $\mathbb R^m$ has jacobian determinant $1$ a.e. Also note that $F^{-1}(\{t\})=\{z \in \mathbb R^m \mid F(z) = t\} = \partial B(t)$ for all $t \ge 0$. By the coarea-formula (see [Corollary 1.4][1], for example), we have
$$
\begin{split}
\int_{B(r)}\mu(z)dV_n(z) &= \int_{0}^r\left(\int_{F^{-1}(\{t\})}\frac{\mu(z)}{|Jac_F(z)|}dS_{m-1}(z)\right)dt\\
&= \int_{0}^r\left(\int_{\partial B(t)}\mu(z)dS_{n-1}(z)\right)dt\\
&= \int_{0}^r\left(\int_{\partial B(t)}\frac{\mu(z)}{S_{m-1}(\partial B(t))}dS_{m-1}(z)\right)S_{n-1}(\partial B(t))dt\\
&\ge\int_{0}^r\left(\int_{\partial B(t)}\frac{\mu(z)^{1/2}}{S_{m-1}(\partial B(t))}dS_{m-1}(z)\right)^2S_{m-1}(\partial B(t))dt\\
&= \int_{0}^r\left(\int_{\partial B(t)}\mu(z)^{1/2}dS_{m-1}(z)\right)^2\frac{1}{S_{m-1}(\partial B(t))}dt,
\end{split}
$$
where the inequality is an applicaiton of Jensen's inequality on the measure probability measure $A \mapsto S_{m-1}(A \cap \partial B(t))/S_{m-1}(\partial B(t))$.

In particular, if $m=2$, then $dS_{m-1} = dS_1$ which is the arc-length measure, and so $S_1(\partial B(t)) =$ length or circlye of radius $t$, which equals $2\pi t$.

  [1]: https://www3.nd.edu/~lnicolae/Coarea.pdf