Solve inequality involving double summation, exponentiation and $\ln$

Question

I have the following inequality (where $n$ is a real number):

$$\sum_{a=1}^n\sum_{b=1}^n\frac{a^n}{n^2}e^{-n-ba}\ge\ln n$$

Computation suggests this holds for $n$ greater than or equal to a number somewhere between $10.75$ and $11$. But can I solve the inequality algebraically? And if so, how?

UPDATE:

I realise that $n$ is strictly speaking only defined for integer values, but rather than just say that the inequality is satisfied (by direct computation) for $n\ge11$, I want to understand the process of solving such an inequality algebraically.

Answer 1

Hint:

\begin{align} \sum_{a=1}^n\sum_{b=1}^n\frac{a^n}{n^2}e^{-n-ba} &=\frac{e^{-n}}{n^2}\sum_{a=1}^n \left(a^n \sum_{b=1}^n (e^{-a})^b\right)\\ &=\frac{e^{-n}}{n^2}\sum_{a=1}^n \dfrac{a^ne^{-a}(e^{-na}-1)}{e^{-a}-1}\\ \end{align}