Show that $45
If $x_0 = 5$ and $x_{n+1} = x_n + \frac {1}{x_n},$ show that
$45<x_{1000}<45.1$
This problem is taken from the list submitted for the $1975$ Canadian Mathematics Olympiad (but not used on the actual exam).
SOURCE : CRUX(Page Number 3 ; Question Number 162)
I tried writing out the first few terms :
$x_1 = 5+ \frac{1}{5} $
$x_2 = \big(5+\frac{1}{5}\big) + \big(5+\frac{1}{5}\big)^{-1} = \frac{x_0^2 + 1}{x_0} +  \frac{x_0}{x_0^2 + 1} = \frac{(x_0^2 + 1)^2+x_0^2}{x_0(x_0^2+1)}$
$x_3 = \frac{(x_0^2 + 1)^2+x_0^2}{x_0(x_0^2+1)} + \frac{x_0(x_0^2+1)}{(x_0^2 + 1)^2+x_0^2} = Messy$
I tried a lot but could not find any general formula for the $n$th term. Does there even exist any?
Also it is clear that $\big(x_n + \frac{1}{x_n}\big)$ is an increasing function. So I think sequence diverges, but how can the $1000th$ term be calculated or aprroximated?
Any help would be gratefully acknowledged :).
 A: CLAIM 
For all $n \ge 1$, then let us prove $$\sqrt{2n+25+\frac{1+\ln (n-1)}{2}} > x_{n} > \sqrt{2n+25}$$
PROOF
This holds for $n=1$. This can be checked numerically. 
Assume this holds for $n=m$. Then, we can prove the left hand inequality through induction as $$x_{m+1}^2=x_{m}^2+2+\frac{1}{x_{m}^2}>2n+27 \implies x_{m+1} > \sqrt{2n+27}$$
Now from $x_{m+1}^2-x_{m}^2=2+\frac{1}{x_{m}^2}$, we now have $$x_{m+1}^2-x_{0}^2=\sum_{n=0}^{m-1} \frac{1}{x_{n}^2}<\sum_{n=0}^{m-1}\frac{1}{2n+25}<\sum_{n=1}^{m}\frac{1}{2n}\le \frac{1+\ln m}{2} \tag{1}$$
Thus, our claim is proved. 
The desired result follows from our Claim. 
$(1)$: See here
NOTE: My original answer was wrong, which is the reason for this rather large edit. 
A: $x_n^2-x_{n-1}^2=2+\frac{1}{x_{n-1}^2}$ for $n\geq1$.
Thus, $$x_{1000}^2=2\cdot1000+25+\frac{1}{x_0^2}+\frac{1}{x_1^2}+...+\frac{1}{x_{999}^2}>2025$$
and $$x_{1000}^2=2\cdot1000+25+\frac{1}{x_0^2}+\frac{1}{x_1^2}+...+\frac{1}{x_{999}^2}<2025+\frac{100}{x_0^2}+\frac{900}{x_{100}^2}<$$
$$<2025+4+\frac{900}{225}=2033<45.1^2$$
Because $x_{100}^2>225$.
