calculus - Convergence of $x_{n+1}=frac{1}{3}(x_n^2+2)$

Let $0

Show that it's bounded above with $x_n <1$

Base Case: $x_1=\frac{1}{3}(x_0^2+2)<1$
Induction Hypothesis: Be $x_k<1$
Induction Step: $n\rightarrow n+1$

$x_{n+1}=\frac{1}{3}(x_n^2+2)<1$.

Show that it's monotonically nondecreasing:
$x_{n+1}-x_n=\frac{1}{3}(x_n^2+2) -x_n=...$
I've made a few steps more, but i can't see why this is in the end $>0$...