Hadamard product operator norm suppose that $A$ and $B$ be $n×n$ nonnegative matrices. Consider the operator (spectral) norm on $A$
$\lVert A \rVert_{op} = sup_{x \ne 0} \dfrac{\lVert Ax \rVert}{\lVert x \rVert}= sup_{\lVert x \rVert=1} \lVert Ax \rVert.$
Then $ \lVert A \circ B \rVert _{op} ≤ \rho (A^{T} B)$?
where $\rho (A)$ spectral radius and $A \circ B=( a_{ij}  b_{ij})$ is a the Hadamard product.
I know that
$\rho (A \circ B ) ≤ \rho (AB),$  and $(A \circ B)(B \circ A) ≤  AB \circ BA$.
 A: Observe that $(A\circ B)^T(A\circ B)\le(A^TB)\circ(B^TA)$ entrywise:
$$
\begin{align}
\big((A\circ B)^T(A\circ B)\big)_{ij}
&=\sum_k\big((A\circ B)^T\big)_{ik}(A\circ B)_{kj}\\
&=\sum_k(A\circ B)_{ki}(A\circ B)_{kj}\\
&=\sum_ka_{ki}b_{ki}a_{kj}b_{kj}\\
&\le\big(\sum_ka_{ki}b_{kj}\big)\big(\sum_kb_{ki}a_{kj}\big)\\
&=(A^TB)_{ij}(B^TA)_{ij}\\
&=\big((A^TB)\circ(B^TA)\big)_{ij}.\tag{0}
\end{align}
$$
It follows that
$$
\|A\circ B\|_2^2
=\rho\big((A\circ B)^T(A\circ B)\big) \le\rho\big((A^TB)\circ(B^TA)\big)
\le\rho(A^TB)\rho(B^TA) =\rho(A^TB)^2.\tag{1}
$$
The first inequality on $(1)$ is true because
$$
0\le X\le Y
\ \Rightarrow\ \forall k\in\mathbb N,\,X^k\le Y^k
\ \Rightarrow\ \rho(X)=\lim_{k\to\infty}\|X^k\|_F^{1/k}\le \lim_{k\to\infty}\|Y^k\|_F^{1/k}=\rho(Y)
$$
by Gelfand's formula. The second inequality is due to the fact when $X$ and $Y$ are square matrices, $X\circ Y$ is a principal submatrix of $X\otimes Y$. Therefore, when $X$ and $Y$ are nonnegative, if $Z$ is the matrix obtained by zeroing out all elements in $X\otimes Y$ outside this principal submatrix, then $0\le Z\le X\otimes Y$ and hence
$$
\rho(X\circ Y)=\rho(Z)\le\rho(X\otimes Y)=\rho(X)\rho(Y).
$$
Update: a literature review reveals that this result was proved using a different argument as corollary 6 in Zejun Huang, On the spectral radius and the spectral norm of Hadamard products of nonnegative matrices, LAA 434 (2011):457-462. A proof along the same line as my answer but presented in a more general manner was given in corollary 2.3 in Dongjun Chen and Fun Zhang, On the spectral radius of Hadamard products of nonnegative matrices, Banach J. Math. Anal. 9(2015), no. 2, 127-133. One interesting observation of the authors is the following inequality for nonnegative matrices:
$$
\begin{aligned}
&(A_{11}\circ A_{21}\circ\cdots\circ A_{s1})(A_{12}\circ A_{22}\circ\cdots\circ A_{s2})\cdots(A_{1t}\circ A_{2t}\circ\cdots\circ A_{st})\\
\le\,&(A_{11}A_{12}\cdots A_{1t})\circ
(A_{21}A_{22}\cdots A_{2t})\circ\cdots\circ
(A_{s1}A_{s2}\cdots A_{st}).
\end{aligned}
$$
In particular, then inequality $(0)$ in my answer can be easily obtained as a special case:
$$
(A\circ B)^T(A\circ B)
=(A^T\circ B^T)(B\circ A)
\le(A^TB)\circ(B^TA).
$$
