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.

Divide by $n$ both sides, you need to prove: $\ln\left(1+x\right)  > \dfrac{2x}{2+x}= 2-\dfrac{4}{2+x}, 0 < x  \le 1, x = \dfrac{1}{n}$. Consider $f(x) = \ln(1+x) +\dfrac{4}{2+x}-2, 0 < x \le 1\implies f'(x) = \dfrac{1}{1+x} - \dfrac{4}{(2+x)^2} = \dfrac{(x+2)^2-4(x+1)}{(1+x)(2+x)^2}= \dfrac{x^2+4x+4-4x-4}{(1+x)(2+x)^2}= \dfrac{x^2}{(1+x)(2+x)^2}> 0\implies f(x) > f(0) = 0\implies \ln(1+x) > 2 - \dfrac{4}{2+x}$. 
