If $(x_n) \to x$ then $(\sqrt[n]{x_1x_2\cdots x_n}) \to x$

By A.M -G.M-H.M inequality we have

$$X_n=\frac{n}{\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n}}\le\{x_1x_2..x_n\}^{\frac{1}{n}} \le Y_n= \frac{x_1+x_2+..x_n}{n}$$..

Now show that both $X_n$ and $Y_n$ converge to $x$ and use Squeeze Principle.

I am going to assume $x_n\geq 0$ for any $n\geq 1$.

Cesàro mean theorem gives that if $\{\alpha_n\}_{n\geq 1}$ is a real sequence converging to $A$ and for every $n\geq 1$ we have $\beta_n=\frac{\alpha_1+\ldots+\alpha_n}{n}$, then $\{\beta_n\}_{n\geq 1}$ is a real sequence converging to $A$, too.

So you just have to consider $\alpha_n=\log x_n$ and derive from $\alpha_n\to \log\alpha$ that $$ \beta_n = \log\sqrt[n]{x_1\cdot x_2\cdots x_n} \to \log\alpha, $$ hence $\sqrt[n]{x_1\cdot x_2\cdots x_n}\to \alpha$ by exponentiating back.