Let A = $\displaystyle [a_{ij}]_{n_x n}$ be a matrix with nonnegative entries such that $\displaystyle \sum\limits^n_{i=1}\sum\limits^n_{j=1}a_{ij}=n$
Prove that $\displaystyle |det\, A|\leq 1$
In $\displaystyle \mathbb{R}^n$, let $\displaystyle z = (1,1,\ldots,1)^{\textsc{t}}$, and let C be the unit cube $\displaystyle [0,1]\times[0,1]\times\cdots\times[0,1]$. Then $\displaystyle AC \subseteq \textstyle \bigl[0,\sum a_{1j}\bigr]\times\Bigl[0,\sum a_{2j}\Bigr]\times\cdots\times\Bigl[0,\sum a_{nj}\Bigr]$.
By the change of variables theorem, with $\displaystyle y = Ax$, the integral of the inner product of y with z over the region AC is given by
$\displaystyle \int_C\langle Ax,z\rangle|\text{det}\,A|\,dx = \int_{AC}\langle y,z\rangle\,dy$.
The left side is $\displaystyle \int_0^1\!\!\!\int_0^1\!\!\!\!\cdots\!\!\!\int_0^1 \Bigl(\sum a_{i1}x_1 + \ldots + \sum a_{in}x_n\Bigr)|\text{det}\,A|\,dx_1\ldots dx_n = \tfrac12\sum_{i,j=1}^n a_{ij}|\text{det}\,A| = \tfrac12|\text{det}\,A|$.
The right side is $\displaystyle \int_{AC}\langle y,z\rangle\,dy \leqslant \int_0^{\sum a_{1j}}\!\!\! \int_0^{\sum a_{2j}}\mkern-30mu\cdots\int_0^{\sum a_{nj}}\!\!\!(y_1+\ldots+y_n)\,dy_1\ldots dy_n = \tfrac12\sum_{i,j=1}^n a_{ij} = \tfrac12.$
Putting the two together, you see that $\displaystyle |\text{det}\,A| \leqslant 1$.
This is a pretty proof, but as such it would actually prove that $\displaystyle |\det A|\leq 1$ for every nonnegative matrix $\displaystyle A$, wouldn't it? I have some concerns about this last line:
One has
$\displaystyle \int_0^{A_1}\cdots\int_0^{A_n}(y_1+\cdots+y_n)dy_1 \cdots dy_n=\frac{1}{2}(A_1)^2 A_2\cdots A_n+\cdots+\frac{1}{2}A_1\cdots A_{n-1}(A_n)^2$ $\displaystyle =\frac{1}{2}A_1\cdots A_n(A_1+\cdots+A_n)$,
which is much larger.
A possible proof: define $\displaystyle \widetilde{A}$ by dividing each line of the matrix $\displaystyle A$ by its sum, hence making it a "stochastic" matrix. We have $\displaystyle \det A=\prod_i\sum_j a_{ij}\det\widetilde{A}$, and by the arithmetic-geometric inequality, the prefactor (which is positive) is less than $\displaystyle \left(\frac{1}{n}\sum_i\sum_j a_{ij}\right)^n\leq 1$, so that we are reduced to proving that $\displaystyle |\det \widetilde{A}|\leq 1$. This is a classical fact, which one can prove by looking at the complex eigenvectors: if $\displaystyle \widetilde{A}Y=\lambda Y$, applying the triangle inequality at the line corresponding to the largest entry $\displaystyle |y_i|$, one gets $\displaystyle |\lambda|\leq 1$. Or there is also a geometrical proof along the exact same lines as yours! since $\displaystyle A_1\cdots A_n=1$ in this case
Yes, that was careless. I was starting from the fact that the linear transformation with matrix A dilates volumes by a factor $\displaystyle |\text{det}\,A|$, and I was looking for a geometric reason why the image of the unit cube under that transformation should have volume at most 1. I still think that method should work, though obviously my approach didn't. However, it stimulated Laurent into providing a correct proof, so some good came of it.
Actually it works in a much simpler way than what you did! (and with the arithmetic-geometric inequality):
We have
$\displaystyle A[0,1]^n\subset \bigg[0,\sum_j a_{1j}\bigg]\times \cdots\times\bigg[0,\sum_j a_{nj}\bigg]$,hence, simply taking the volume of each side:
$\displaystyle |\det A|\leq \prod_i \sum_j a_{ij}$.By the arithmetic-geometric inequality, the right-hand side is less than $\displaystyle \left(\frac{1}{n}\sum_{i,j}a_{ij}\right)^{n}$. Under the assumption of the question, this equals 1. qed