<s>[INST] A simple proof that $\bigl(1+\frac1n\bigr)^n\leq3-\frac1n$? The inequality
$$
e_n:=\left(1+\frac1n\right)^n\leq3-\frac1n,
$$
where $n\in\mathbb{N}_+$, is certainly true, because we know, how LHS is connected with $e$. The other argument is the standard proof of boundedness of $(e_n)$, which uses the binomial theorem.
Are there any more elementary proofs of this inequality?
 [/INST] You can use induction. 
For $n=1$, clearly $2 \leq 2$. Assume it holds for some $n-1 \in \mathbb{N}$. Then,
$$\left(1+ \frac{1}{n} \right)^n = \left( 1 + \frac{1}{n} \right)^{n-1} \left( 1 + \frac{1}{n} \right) \leq \left( 1 + \frac{1}{n-1} \right)^{n-1} \left( 1 + \frac{1}{n} \right) \leq \left( 3- \frac{1}{n-1} \right) \left(1 - \frac{1}{n} \right) = 3 - \frac{3}{n} - \frac{1}{n-1} + \frac{1}{n(n-1)}.$$
It is left to show
$$-\frac{3}{n} - \frac{1}{n-1} + \frac{1}{n(n-1)} \leq -\frac{1}{n}$$
or equivalently
$$ -3n+3-n+1 \leq -n+1,$$
i.e. $n \geq 1$. Since $n-1 \in \mathbb{N}$, we have $n \geq 2$ by choice, so the inequality holds.
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