@NoahSchweber ah okay. The interesting question is about maximality anyway

@NoahSchweber if $A$ is precoded then that conflicts with maximality does it not?

@NoahSchweber but how do you make it above $A$?

I see you also used that by L'Hôpital's rule, \begin{eqnarray*} \lim_{x\to\infty} x^2(\ln x-\ln(x-1))-x &=& \lim_{x\to\infty} \frac{\ln x- \ln(x-1)-1/x}{1/x^2} \\ &=&\lim_{x\to\infty} \left(\frac1x- \frac1{x-1}+\frac1{x^2} \right)\bigg/\left(-2/x^3\right)\\ &=&\lim_{x\to\infty} \left(\frac{1}{x^2(x-1)} \right)\bigg/\left(2/x^3\right)=‌​1/2. \end{eqnarray*}

Thanks! These are rather specific integers but still, good to know.

Great... much obliged

Right, right. In terms of $c$ and in terms of an $n$ with all $a_i<n$.

Thanks, I didn't think of using AM-GM here, how are you doing that exactly?

Much obliged. Paper looks non-elementary :)