Suppose $A$ is an $n\times n$ stochastic matrix, that is, entrywise nonnegative and row sums are all $1$. If $A$ is invertible, is it true that the minimum diagonal entry of $A^{-1}$ is no larger than $1$? 


Small matrices support this claim, but for larger ones, I don't know how to (dis)prove it. 


**Edited** I forgot to add the condition that the diagonal entries of $A$ are all zero.



Bounding the minimum entry of an inverse matrix
Here's a counterexample:
$$ A = \frac{1}{5}
 \left[ \matrix{ 0 & 1 & 3 & 1 \cr 2 & 0 & 1 & 2 \cr 3 & 1 & 0 & 1 \cr 0 & 3 & 2 & 0}\right], \quad
 A^{-1}= \frac{1}{3} \left[ \matrix{ 7 & -9 & 11 & -6 \cr -8 & 6 & -4 & 9 \cr 12 & -9 & 6 & -6 \cr -13 & 21 & -14 &9} \right].
$$


