Is $\det(I+A)\le 1$ if $A$ is a nonnegative matrix with zero diagonals and all row sums less than 1?
I am inclined to say yes. Note by Gershgorin's circle theorem all eigenvalues of $A$ has absolute value less than 1, so we have $\det(I+A)=\exp (\sum_{j=1}^\infty (-1)^{j-1}\frac{\mathrm{tr}(A^j)}{j})$. The $j=1$ term is $0$ since $\mathrm{tr}(A)=0$, and the second order term corresponding to $j=2$ is negative, and this should be the dominant term? If this is not true, what if we further impose the column sums are less than 1?
Edit: Thanks for the counterexample. Bonus question: is there any uniform bound on the entries that can make the assertion true, say 1/4? (this question actually arises in a graph theory problem, and the matrix $A$ has something to do with the squares of the reciprocal of the degree sequence, so in this case entries can have than 1/4)
As a counterexample, if $$A = \begin{bmatrix}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{bmatrix},$$ then $\det(I+A)=2$.
If row sums should be strictly less than $1$, then take $(1-\varepsilon)A$ instead, for a determinant of $1 + (1-\varepsilon)^3 = 2 - O(\varepsilon)$.
For the bonus question: If we take the matrix $\delta A$ for any $\delta > 0$, then we have a matrix whose row sums, column sums, and entries are all bounded by $\delta$, but the determinant $\det(I + \delta A)$ is still bigger than $1$: more precisely, it is $1 + \delta^3$. (Optimizing numerically for some values of $\delta$, it seems like this is the best we can do for $3\times 3$ matrices, and conjecturally, it might be the best we can do in general.)