How to show that: if $n\ln\left(1+a/n\right)\geqslant k\ln\left(1+a/k\right)$ then $n\geqslant k$? Let $a>0$ and $n,k$ positive integers.
If 
$$n\ln\left(1+a/n\right)\geqslant k\ln\left(1+a/k\right),$$
then
$$n\geqslant k.$$
I tried by contrapositive by I do not get much. If $n<k$ then I would have 
$$\ln\left(1+a/n\right)>\ln\left(1+a/k\right),$$
which does not help me a lot.

Let
$f(x)
=x \ln(1+a/x)
$.
$\begin{array}\\
f'(x)
&=\ln(1+a/x)+x(\ln(1+a/x))'\\
&=\ln(1+a/x)+x\frac{(1+a/x)'}{1+a/x}\\
&=\ln((x+a)/x)+x\frac{-a/x^2}{1+a/x}\\
&=-\ln(x/(x+a))-\frac{a}{x+a}\\
&=-\ln(1-a/(x+a))-\frac{a}{x+a}\\
&\gt a/(x+a)-\frac{a}{x+a}
\qquad\text{since }-\ln(1-z) > z \text{ for }z > 0\\
&= 0\\
\end{array}
$
Therefore
$f(x)$
is increasing,
which is what you want.
