Consider the matrices $m\times m$ $K,L,M$.
The symbol $\circ$ denotes term by term product. The subscript $++$, as in $K_{++}$, denotes the sum of all the terms of the matrix.
$\boldsymbol{1}_m$ is the $m\times 1$ vector of ones.
Could you help me to show that $$ \operatorname{trace}\Big((\boldsymbol{1}_m \boldsymbol{1}_m'K) \circ (\boldsymbol{1}_m \boldsymbol{1}_m'L) \circ (\boldsymbol{1}_m \boldsymbol{1}_m'M)\Big)=(KLM)_{++} $$
What I have shown so far is that $$ \operatorname{trace} \Big((\boldsymbol{1}_m \boldsymbol{1}_m'K) \circ (\boldsymbol{1}_m \boldsymbol{1}_m'L) \circ (\boldsymbol{1}_m \boldsymbol{1}_m'M)\Big) $$ $$=\sum_{i=1}^mK_{i,1}\sum_{i=1}^mL_{i,1} \sum_{i=1}^m M_{i,1}+\dots +\sum_{i=1}^mK_{i,n}\sum_{i=1}^m L_{i,n}\sum_{i=1}^m M_{i,n} $$ and $$ (KLM)_{++}=\sum_{g=1}^m\Big(\sum_{j=1}^n M_{j,g}\sum_{i=1}^n K_{1,i}L_{i,g}\Big)+...+\sum_{g=1}^m\Big(\sum_{j=1}^n M_{j,g}\sum_{i=1}^n K_{n,i}L_{i,g}\Big) $$ but Iam struggling to go ahead.
Since the wanted identity is linear in $K$, $L$ and $M$, it suffices to treat the case where $K=E_{i,j}$, $L=E_{i',j'}$ and $M=E_{i'',j''}$, where $E_{i,j}$ denotes the matrice whose $(i,j)$ entry is $0$ and all the others are zero.
Observe that $ \mathbf 1_m\mathbf 1'_mE_{i,j}=\sum_{u=1}^mE_{u,j}$. For the last term, use the fact that $$E_{i,j}E_{i',j'}=\begin{cases} E_{i,j'}&\mbox{ if } j=i'\\ 0&\mbox{ otherwise}. \end{cases} $$