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