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$$ 1+\frac{2}{3n-2}\leqslant \sqrt[n]{3}\leqslant 1+\frac{2}{n}, n\in \mathbb{Z}^{+} $$

How to prove this inequation?

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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

$$\sqrt[n]{3} \le \frac{1}{n}\left( (n-1)\times 1 + 3 \right) = 1 + \frac{2}{n}$$

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The right hand side is Binomial theorem for we have $$(1+\frac{2}{n})^{n}=1+n.\frac{2}{n}+{n \choose 2}(\frac{2}{n})^{2}+\cdots \geq1+2=3$$ and hence $3^{\frac{1}{n}}\leq (1+\frac{2}{n})$, with equality holding only when $n=1$.(Since the $n$-th root fuction is increasing).

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The left inequality (rather both) is straightforward using calculus. Consider $f:(\frac{2}{3}, \infty) \to \mathbb{R}$ $$f(x)=3-(1+ \frac{2}{3-2x})^x$$

Clearly $f(1) = 0$. Differentiating, $$f'(x)=6 \left( 1+\frac{3}{3x-2}\right)^x \log \left( 1+\frac{2}{3x-2}\right)\frac{1}{(3x-2)^2}$$

which is positive for $x \geq 1$. Hence the result.

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