TITLE: Prove that $\left(\left\{\sqrt{x^4+18x^2+162}\right\}\right)$ is a strictly decreasing sequence
QUESTION [5 upvotes]: Prove that $\left(\left\{\sqrt{x^4+18x^2+162}\right\}\right)$ is a strictly decreasing sequence where $x \in \mathbb{Z}^+$ after some point. Note: $\{r\}$ denotes the fractional part of a real number $r$. 

In order to show this, we must show that for all $x$ we have $$\left\{\sqrt{x^4+18x^2+162}\right\} > \left\{\sqrt{(x+1)^4+18(x+1)^2+162}\right\}.$$ How do we show this?

REPLY [9 votes]: Step 1. Prove
$$x^2+9<\sqrt{(x^2+9)^2+9^2}<x^2+10$$
Left inequality is obvious, right side inequality is also true for $x>5$ which is also trivial to show.
So you have $\lfloor\sqrt{(x^2+9)^2+9^2}\rfloor=x^2+9$
Step 2. Write:
$$\left\{\sqrt{(x^2+9)^2+9^2}\right\}=\sqrt{(x^2+9)^2+9^2}-\sqrt{(x^2+9)^2}=\frac{9^2}{\sqrt{(x^2+9)^2+9^2}+\sqrt{(x^2+9)^2}}$$
which is decreasing
The first 6 terms can be shown to be decreasing by direct computation.
EDIT: I checked by direct computation, and the first 6 terms of the sequence do not decrease, but the sequence decreases after the 6th term, as proved above
0.45362404707370985
0.8113883008418963
0.12461179749810825
0.5706605111728464
0.17101079013794873
0.8911756223350622
0.6941223633167581
0.5527021937331682
0.4488805900880095
0.3709284956473198