Prove that if $\lim_{n\to\infty} x_n = 1$ for $x_n > 0$ then $\lim_{n\to\infty} \sqrt[n]{x_1x_2\cdots x_n} = 1$ 
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?
 A: 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.
A: Consider the sequence defined, for all $n \geq 0$, by$$y_n = \ln(x_n)$$
Obviously $y_n \rightarrow 0$. So by the Cesaro theorem, you have that 
$$\frac{y_1 + ... + y_n}{n} \rightarrow 0$$
That means that 
$$\frac{\ln(x_1 \times ... \times x_n)}{n} \rightarrow 0$$
i.e. 
$$\ln \left( \sqrt[n]{x_1 \times ... \times x_n} \right) \rightarrow 0$$
You deduce that 
$$ \sqrt[n]{x_1 \times ... \times x_n} \rightarrow 1$$
A: Just for the completeness: we can use the $HM\le GM\le AM$ inequality:
$$
H_n:=\frac{1}{\frac{\sum_{k=1}^n {\frac{1}{x_k}}}{n}} \le
\left(\prod_{k=1}^n {x_k}\right)^{\frac{1}{n}}\le 
\frac{\sum_{k=1}^n {x_k}}{n}=:A_n
$$

*

*$x_n\to 1 \text{ with } x_n>0 \implies \frac{1}{x_n}\to 1$

*Cesaro summation Cauchy's first limit theorem implies that $\frac{1}{H_n}\to 1$ (so $H_n\to 1$ ) and $A_n\to 1$

*apply the Squeeze theorem
