A little question about convergence of sequence

Question

It's known that: $$\begin{cases}x_n=\sqrt[3]{6+x_{n-1}}\\x_1 = \sqrt[3]{6}\end{cases}$$ $$x_n\to2,\space x_n\uparrow,\space x_n\in(0, 2)$$ $$a_n=\frac{x_n^2+2x_n+4}{12}\space, a_n\to1,\space a_n\uparrow,\space a_n\in(0,1)$$ $$a_1\approx0.91,\space a_2\approx0.992,\space a_3\approx0.9994,\space a_4\approx0.99995, ...$$ $$p_n=a_1a_2a_3...a_n,\space p_n\downarrow,\space p_n\in(0,1)$$

How to prove that $p_n\not\to0$?

I believe that there should be some lower bound greater than $0$ but how to find it i don't know.

Thanks!

Answer 1

Hint. $x_{n-1}-2 = x_n^3-8 = (x_n-2)(x_n^2+2x_n+4) = 12a_n(x_n-2)$.