Is the spectral radius of $DA$ less than the spectral radius of $A$ when $D$ is diagonal where all diagonal entries are nonnegative and less than 1?
This is true when $A$ is normal, since $$ \rho(DA) \le \|DA\|\le \|D\| \| A\| \le \|A\| = \rho(A) $$
My guess is that it is false in general.
Notice that it is enough to prove $\|(DA)^k\|\le \|A^k\|$ definitively in $k$.
If we let $D$ have negative values, then it would imply that any sign change in any row does not change the spectral radius, that is preposterous.