The spectral radius is monotone on nonnegative matrices

Entrywise inequality for powers. We prove $0\le A^k\le B^k$ entrywise by induction. For $k=1$ it is the hypothesis. If $0\le A^k\le B^k$, then
$$(A^{k+1})_{ij}=\sum_l(A^k)_{il}A_{lj}\le\sum_l(B^k)_{il}B_{lj}=(B^{k+1})_{ij},$$
because all terms are nonnegative and each factor is dominated.

The norm. For a matrix with nonnegative entries, $\|M\|_\infty=\max_i\sum_jM_{ij}$ is nondecreasing in the entries. Hence
$$\|A^k\|_\infty\le\|B^k\|_\infty\quad\text{for all }k.$$

Gelfand. Taking $k$-th roots and letting $k\to\infty$ (Gelfand's formula for the submultiplicative norm $\|\cdot\|_\infty$):
$$\rho(A)=\lim_k\|A^k\|_\infty^{1/k}\le\lim_k\|B^k\|_\infty^{1/k}=\rho(B).$$

Example. $A=\begin{pmatrix}0&1\\0&0\end{pmatrix}\le B=\begin{pmatrix}1&1\\1&0\end{pmatrix}$: $\rho(A)=0\le\rho(B)=\frac{1+\sqrt5}2$.

Remark. Nonnegativity is essential: for $A=\begin{pmatrix}0&-3\\0&0\end{pmatrix}\le B=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ (entrywise) there is no problem, but with $A=\begin{pmatrix}-2&0\\0&-2\end{pmatrix}\le B=I$ we get $\rho(A)=2>\rho(B)=1$. This monotonicity is a first step of the Perron-Frobenius theory.