Proof for $e^{\frac{1}{n+1}}-1-\frac{1}{n}\leq0$ I'm looking for a proof of $e^{\frac{1}{n+1}}-1-\frac{1}{n}\leq0$, optionally $\ln(n)+\frac{1}{n+1}\leq \ln(n+1)$

If $x \ge 0$, 
$$e^x = 1 + \int_0^x e^t\, dt \le 1 + \int_0^x e^x\, dt = 1 + xe^x.$$
So $(1 - x)e^x \le 1$, and thus 
$$e^x \le \frac{1}{1 - x},\quad 0 \le x < 1.$$
Letting $x = \frac{1}{n+1}$, we obtain
$$e^{\frac{1}{n+1}} \le \frac{1}{1 - \frac{1}{n+1}} = \frac{n+1}{(n+1)-1} = \frac{n+1}{n} = 1 + \frac{1}{n}.$$
Equivalently, 
$$e^{\frac{1}{n+1}} - 1 - \frac{1}{n} \le 0.$$
