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.
 A: Let $\dfrac{4-t}{t}$ be $1+\dfrac{1}{n}$. Then one has
$$\ln\left(1+\dfrac{1}{n}\right)=\ln\left(\dfrac{4-t}{t}\right)\ge 2-t=\dfrac{2}{2n+2}.$$
Multiply both sides by $n$ and the result follows.
A: Your inequality is a direct consequence of Jensen's inequality in integral form


*

*$\frac 1{b-a}\int_a^b \phi\left(f(x)\right)\;dx \geq \phi\left(\frac 1{b-a}\int_a^b f(x)\;dx\right)$ setting

*$\phi(t) = \frac 1{1+t}$, $f(x)=x, a=0, b=\frac  1n$
$$n\ln\left(1+\frac 1n\right)= n \int_0^{\frac 1n}\frac{dx}{1+x}\geq \frac 1{1+n\int_0^{\frac 1n}x\;dx}=\frac 1{1+\frac 1{2n}}=\frac{2n}{2n+1}$$
A: 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 $f(x) = \ln(1+x) +\dfrac{4}{2+x}-2, 0 < x \le 1\implies f'(x) = \dfrac{1}{1+x} - \dfrac{4}{(2+x)^2} = \dfrac{(x+2)^2-4(x+1)}{(1+x)(2+x)^2}= \dfrac{x^2+4x+4-4x-4}{(1+x)(2+x)^2}= \dfrac{x^2}{(1+x)(2+x)^2}> 0\implies f(x) > f(0) = 0\implies \ln(1+x) > 2 - \dfrac{4}{2+x}$. 
A: 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}
$.
