For some vector $v$ in the simplex (i.e., $v_k \geq 0, \forall k$ and $\sum_k v_k = 1$) and a matrix $A$ with all entries strictly positive, consider $$J(v) = J_1(v) + J_2(v),$$ with $$J_1(v) = (\mathrm{diag}(A v))^{-1} A \mathrm{diag}(v)$$ and $$ J_2(v) = (\mathrm{diag}(A v))^{-1} A \mathrm{diag}(A v)].$$
I am interested in understanding how the spectral radius of $J(v)$ depends on $v$.
Note that $J_1(v)$ is a stochastic matrix and hence its spectral radius is one, while the spectral radius of $J_2(v)$ is equal to the spectral radius of $A$. This implies that, for both $J_1(v)$ and $J_2(v)$, the spectral radius is independent of $v$. Of course, this is no longer the case for $J(v)$.
Note also that if $A$ is a diagonal matrix then $J_1(v) = I$ and $J_2(v) = A$, implying that $J(v) = I + A$ is independent of $v$ and so the spectral radius of $J(v)$ is weakly maximized at any corner of the simplex, and it is equal to 1 plus the spectral radius of $A$. In the case of $A$ being lower or upper triangular, $J(v)$ is no longer independent of $v$, but one can easily see that its spectral radius is maximized at a corner and it is also 1 plus the spectral radius of $A$.
Simulations suggest that the spectral radius of $J(v)$, or $\rho(J(v)$, always attains its maximum at a corner of the simplex. One idea I had was that $\rho(J(v))$ would be convex in $v$, but simulations show that this is not true. What is true in general, according to simulations, is that there is no internal maximum of $\rho(J(v))$.
Any ideas for how to prove that $\rho(J(v))$ is maximized at the corner of the simplex would be much appreciated.