real analysis - Prove that if $lim_{ntoinfty} x_n = 1$ for $x_n > 0$ then $lim_{ntoinfty} sqrt[n]{x_1x_2cdots x_n} = 1$

Given a sequence $x_n$ and the fact that:

$$\lim_{n\to\infty} x_n = 1\\ x_n > 0\\ n\in\Bbb N$$
Prove
$$\lim_{n\to\infty} \sqrt[n]{x_1x_2\cdots x_n} = 1$$

I'm having some difficulties finishing this proof. I've shown while solving another problem that:

$$\lim_{n\to\infty} x_n = a \implies \lim_{n\to\infty}{1\over n}\sum_{k=1}^nx_k = a$$

Using this we may state that:

$$\lim_{n\to\infty}x_n = 1 \implies \lim_{n\to\infty}{1\over n}\sum_{k=1}^nx_k = 1$$

On the other hand by AM-GM we have that:

$$\frac{x_1 + x_2 + \cdots x_n}{n} \ge \sqrt[n]{x_1x_2\cdots x_n}$$

Since $x_n > 0$:

$$\frac{x_1 + x_2 + \cdots x_n}{n} \ge \sqrt[n]{x_1x_2\cdots x_n} \ge 0$$

We know that:

$$\lim_{n\to\infty}\frac{x_1 + x_2 + \cdots x_n}{n} = 1$$

Therefore:

$$1 \ge \lim_{n\to\infty}\sqrt[n]{x_1x_2\cdots x_n} \ge 0$$

My idea was to use Monotone Convergence theorem, but since $x_n$ is only constrained by $x_n > 0$ we can not make any conclusions on the monotonicity of:

$$y_n = \sqrt[n]{x_1x_2\cdots x_n}$$
(or can we?).

Apparently my idea to use MCT is not applicable here. So the question is what would be the proper way to prove the above?

Answer

Consider the logarithm and use Stolz cesaro to deduce that

$$\lim_{n\to \infty}\log\sqrt[n]{x_1\dotsb x_n}=\lim_{n\to \infty}\frac{1}{n}\sum_{i=1}^n\log x_i=\lim_{n\to \infty}\log x_n=0$$
since $x_n\to 1$ from which the claim follows.