<s>[INST] How to prove $\ln{\frac{n+1}{n}}\le\frac{2}{n+1},\forall n\in\mathbb{N}^+$? I found an inequality: $$\ln\left(\frac{n+1}{n}\right)\le\frac{2}{n+1},\forall n\in\mathbb{N}^+.$$
I tried induction. It is obvious if $k=1$, when $n=k$, $\ln\sqrt{\left(\frac{k+1}{k}\right)^{k+1}}\le 1$, but bogged down for $n=k+1$:
$$\ln\sqrt{\left(\frac{k+2}{k+1}\right)^{k+2}}\le 1$$
 [/INST] Your question is equivalent to prove the following form
$$
(2-\frac{n-1}{n})\leq 2^{(1-\frac{n-1}{n+1})} \tag{1}
$$
The relation $(1)$ is true becuse $\frac{n-1}{n} \geq \frac{n-1}{n+1}$. 
Edit:
If the relation $(1)$ be true then we can conclude that:
$$
\left\{
\begin{array}{c}
(2-\frac{n-1}{n})\leq 2^{(1-\frac{n-1}{n+1})} \\
\\
2^{(1-\frac{n-1}{n+1})}  \leq e^{(1-\frac{n-1}{n+1})} 
\end{array}
\right.
\Rightarrow 
(2-\frac{n-1}{n})\leq e^{(1-\frac{n-1}{n+1})} \tag{2}
$$
$$
\ln (2-\frac{n-1}{n})\leq (1-\frac{n-1}{n+1})
\Rightarrow  
\ln\left(\frac{n+1}{n}\right)\le\frac{2}{n+1}
$$
</s>