Show that $n \ln \left(1+\frac{1}{n}\right) \geq \frac{2 n}{2 n+1}$ 
Hence or otherwise show that for all positive integers $n$
$$
n \ln \left(1+\frac{1}{n}\right) \geq \frac{2 n}{2 n+1}
$$

This is related to another part of this question where I proved $\ln \left(\frac{4-t}{t}\right) \geq 2-t$ for $0<t \leq 2$ by integrating $\frac{1}{s}+\frac{1}{4-s} \geq 1$(which is true for $0<s<4$) over $[t, 2] .$ However, I am not able to use this to prove that
$$n \ln \left(1+\frac{1}{n}\right) \geq \frac{2 n}{2 n+1}$$ for all positive integers. Any help would be appreciated.

Integrating
$\dfrac1{1+t}
=\sum_{k=0}^{m-1}(-1)^k t^k+\dfrac{(-1)^mt^m}{1+t}
$,
we get
$\ln(1+x)
=\sum_{k=0}^{m-1}(-1)^k \dfrac{x^{k+1}}{k+1}+\int_0^x \dfrac{(-1)^mt^m}{1+t}dt
$
for any $m \ge 1$ and $x \ge 0$.
Therefore
$n\ln(1+1/n)
=\sum_{k=0}^{m-1}(-1)^k \dfrac1{(k+1)n^k}+n\int_0^{1/n} \dfrac{(-1)^mt^m}{1+t}dt
$.
Putting
$m = 2$, this is
$n\ln(1+1/n)
=1-\dfrac1{2n}+n\int_0^{1/n} \dfrac{t^2}{1+t}dt
=\dfrac{2n-1}{2n}+n\int_0^{1/n} \dfrac{t^2}{1+t}dt
$
so we want
$\dfrac{2n}{2n+1}
\le \dfrac{2n-1}{2n}+n\int_0^{1/n} \dfrac{t^2}{1+t}dt
$
or
$\dfrac{1}{2n^2(2n+1)}
\le \int_0^{1/n} \dfrac{t^2}{1+t}dt
$.
$\int_0^{1/n} \dfrac{t^2}{1+t}dt
\ge \int_0^{1/n} \dfrac{t^2}{1+1/n}dt
=\dfrac{(1/n)^3}{3(1+1/n)}
=\dfrac1{3n^2(n+1)}
$
so we want
$3n^2(n+1)
\le 2n^2(2n+1)
$
or
$3(n+1)
\le 2(2n+1)
$
or
$3n+3 \le 4n+2
$
or
$n \ge 1$
which is true.
Also,
$\int_0^{1/n} \dfrac{t^2}{1+t}dt
\lt \int_0^{1/n} t^2dt
=\dfrac1{3n^3}
$
so
$n\ln(1+1/n)
=1-\dfrac1{2n}+n\int_0^{1/n} \dfrac{t^2}{1+t}dt
\lt 1-\dfrac1{2n}+\dfrac1{3n^2}
\lt 1-\dfrac{3n-2}{6n^2}
$.
