How to prove this inequation? $$
1+\frac{2}{3n-2}\leqslant \sqrt[n]{3}\leqslant 1+\frac{2}{n}, n\in \mathbb{Z}^{+}
$$
How to prove this inequation?

For the LHS, apply $\text{AM} \ge \text{GM}$ to $n$ numbers with $n-1$ copies of $1$ and one copy of $\frac13$, we get
$$\begin{align}
&\left(1 + \frac{2}{3n-2}\right)^{-1}= 1 - \frac{2}{3n} = \frac{1}{n}\left( (n-1)\times 1 + \frac{1}{3}\right) \ge \frac{1}{\sqrt[n]{3}}\\
\implies &
1 + \frac{2}{3n-2}  \le \sqrt[n]{3}
\end{align}$$
For the RHS, apply $\text{AM} \ge \text{GM}$ to $n$ numbers with $n-1$ copies of $1$ and one copy of $3$, we get
$$\sqrt[n]{3} \le \frac{1}{n}\left( (n-1)\times 1 + 3 \right) = 1 + \frac{2}{n}$$
