TITLE: How to prove this inequality $x^2_{n}\le\frac{8}{3}$
QUESTION [23 upvotes]: let sequence $\{x_{n}\}$ such $x_{1}=0,x_{2}=1$,and 
$$x_{n+1}=\left(1+\dfrac{1}{n}\right)x_{n}-x_{n-1},n\ge 2$$
show that
$$x^2_{n}\le\dfrac{8}{3}$$
This problem it seem interesting,and 
$$x_{n+1}-x_{n}=\dfrac{1}{n}x_{n}-x_{n-1}$$
so we have
$$x_{n+1}-x_{1}=\sum_{k=1}^{n}\dfrac{x_{k}}{k}-\sum_{k=1}^{n-1}x_{k}$$
 where $x_{0}=-1$
It seem this problem very interesting,I guess this $\dfrac{8}{3}$ maybe is not best constant,But is stronger constant

REPLY [12 votes]: Just so this question is fully answered (i.e. getting the bound $8/3$) here I add a condensed proof of $x_n^2 \leq \frac{8}{3}$ from the answer given here as Community Wiki.

Define 
$$E_n = (x_n - x_{n-1})^2 + F_n\left(x_n + x_{n-1}\right)^2$$ 
with $F_n = \frac{(n-1)}{3n+1}$. The recurrence relation gives us
$$E_{n+1} - E_n = \frac{(F_{n+1}-F_n)}{F_n}\cdot F_n\left(x_n + x_{n+1}\right)^2 \geq 0$$
since $F_n$ is a positive and increasing function. This shows that $E_n$ is increasing. We also see that
$$E_{n+1} - E_n \leq \frac{(F_{n+1}-F_n)}{F_n}\cdot E_n \implies \frac{E_{n+1}}{F_{n+1}} \leq \frac{E_n}{F_n}$$
so $\frac{E_n}{F_n}$ is decreasing and since $F_n$ converges it follows that $E_n$ converges. To bound $x_n$ from $E_n$ note that the definition of $E_n$ (the equation for an ellipse) allows us to write $x_n - x_{n-1} = \sqrt{E_n}\cos(\phi_n)$ and $x_n + x_{n-1} = \sqrt{\frac{E_n}{F_n}}\sin(\phi_n)$ for some angle $\phi_n$. Thus
$$x_n^2 = \frac{E_n}{4}[\cos(\phi_n) + \frac{1}{\sqrt{F_n}}\sin(\phi_n)]^2$$
Maximizing the right hand side over $\phi_n$ gives us
$$x_n^2 \leq \frac{E_n}{4}\left[1 + \frac{1}{F_n}\right] = E_n\cdot \frac{n}{n-1}$$
Since $\frac{E_n}{F_n}$ is decreasing we have $E_n \leq F_n \frac{E_2}{F_2} = 8F_n$ and
$$x_n^2 \leq \frac{8n}{3n+1} \leq \frac{8}{3}$$
It is possible to improve the constant with a bit of numerical computation (using $n=5$ instead of $n=2$ when bounding $E_n$) to the best possible one $\frac{9}{4}$ (for which we have equality for $n=3$). The best possible asymptotic bound would be slightly smaller than $2$ (best possible in the sense that $x_n^2 \approx 2$ would occur infinitely many times).