For bounded linear operators $A$ and $B$ on a Banach space $X$, I'm looking for results which imply that $r(A) < r(A+B)$ (note the strict inequality), where $r(A)$ denotes the spectral radius of $A$. Are there any assumptions on the operators $A$, $B$ which will guarantee this?

I should not have left this so general; specifically, I have two integral operators $A$, $B$ on $L^1(0,1)$. I know the spectral radius for $A$, but all I know about $B$ is that it's compact with a non-negative kernel. Ideally, I would like to show that $r(A)<r(A + B)$.

Here's some more information regarding the operators: $$(A\psi)(y) := \int_0^1 s(x)g(y,x) \psi(x) \, dx,$$ and $$(B \psi)(y) := \int_0^1 d(y)f(x) \psi(x) \, dx$$ where all component functions are continuous and non-negative (in fact, take $s(x)$, $d(y)$, and $f(x)$ to be strictly positive on $[0,1]$). The function $g(y,x)$ has some properties that allowed me to compute $r(A)$ exactly.

These operators came up in a mathematical ecology model, and none of them have self-adjoint kernels.