I want to prove the following: Let $\{\sigma_{(m,k)}\}_{m,k}$ be a family of positive real values with $m=1,\ldots,NL$ and $k=1,\ldots,K$ where $N,L,K\in\mathbb{N}$. This family satisfies that $\sigma_{(m+1,k)}\leq \sigma_{(m,k)}$ for all $k$ and $m$. Then,
\begin{equation} \sum_{k=1}^{K}\sum_{m=L+1}^{NL}\sigma_{(m,k)}^2\leq\frac{1}{LK}\left(\sum_{k=1}^K\sum_{m=1}^{NL}\sigma_{(m,k)}\right)^2 \end{equation}
It is possible to prove the case $K=1$ because you can define the family of index $\{I_n\}$ such that $I_n = \{m\in\mathbb{N}:1+(n-1)L\leq m\leq1+nL\}$. Then you have that for all $n=1,..,N/L-1$, \begin{equation} \sigma_{(\hat{m})}\leq\frac{1}{L}\sum_{m\in I_n}\sigma_{(m)}\qquad \forall\hat{m}\in I_{n+1} \qquad(\dagger) \end{equation}
Then, using $(\dagger)$ you have
\begin{align} \sum_{k=1}^{K}\sum_{m=L+1}^{NL}\sigma_{(m)}^2&=\sum_{k=1}^{K}\sum_{n=2}^{N/L}\sum_{m\in I_n}\sigma_{(m)}^2\leq\sum_{k=1}^{K}\sum_{n=2}^{N/L}\sum_{m\in I_n}\left(\frac{1}{L}\sum_{m\in I_{n-1}}\sigma_{(m)}\right)^2\\ &\leq\leq\sum_{k=1}^{K}\sum_{n=2}^{N/L}\frac{1}{L}\left(\sum_{m\in I_{n-1}}\sigma_{(m)}\right)^2\leq\frac{1}{L}\left(\sum_{k=1}^K\sum_{m=1}^{NL}\sigma_{(m)}\right)^2 \end{align}
But I have not been able to prove the general case.
This is not true. Suppose $\sigma_{(m,1)}=1$ for every $m$ and $\sigma_{(m,k)}=\epsilon>0$ for every $k>1$. Then \begin{aligned} A&=\sum_{k=1}^{K}\sum_{m=L+1}^{NL}\sigma_{(m,k)}^2 =(N-1)L\left[1+(K-1)\epsilon^2\right],\\ B&=\frac{1}{LK}\left(\sum_{k=1}^K\sum_{m=1}^{NL}\sigma_{(m,k)}\right)^2 =\frac{N^2L}{K}\left[1+(K-1)\epsilon\right]^2. \end{aligned} Therefore $A>B$ when $\epsilon$ is sufficiently small and $(N-1)K>N^2$.