Let $(c_{nr})$ be an $N\times R$ complex matrix, then $\forall z_n \in \mathbb{C}$, we have $$ \sum_r \Big\sum_n c_{nr}z_n\Big^2 \geq \frac{1}{\sigma_{max}} \sum_n z_n^2 $$ where $\sigma_{max}$ is the maximal sigular value of the complex matrix $(c_{nr})$.

2$\begingroup$ This is false: take for $(z_n)$ a nonzero vector in the kernel of the matrix. $\endgroup$ – abx Sep 16 '20 at 12:16

2$\begingroup$ the correct inequality should have $\sigma_{\rm min}^2$ instead of $1/\sigma_{\rm max}$ $\endgroup$ – Carlo Beenakker Sep 16 '20 at 12:18

2$\begingroup$ This cannot be true on dimensional grounds (imagine scaling the matrix by a factor $\lambda$). $\endgroup$ – gmvh Sep 16 '20 at 12:24
The $N\times R$ matrix $C$ has elements $c_{nr}$, the $N\times N$ matrix $Z$ has elements $z_n \bar{z}_m$, and $C^\ast$ is the conjugate transpose of $C$. The Hermitian matrix product $CC^\ast$ has eigenvalues $\sigma_n^2$, with $\sigma_n\geq 0$, $n=1,2,\ldots N$ the set of singular values of $C$. Then we have $$\sum_r \left\sum_n c_{nr}z_n\right^2 = {\rm tr}\, (CC^\ast Z) = \sum_{n=1}^N \sigma_n^2 \zeta_n^2\geq \sigma_{\rm min}^2 \sum_n \zeta_n^2=\sigma_{\rm min}^2 \sum_n z_n^2,$$ with $\sigma_{\rm min}$ the smallest of the singular values and the vector $\zeta$ obtained from $z$ by a unitary transformation. This is not the inequality in the OP, which should have $\sigma_{\rm min}^2$ instead of $1/\sigma_{\rm max}$.