Do there exist two bounded sequence such that $a_m-a_n> \frac{1}{\sqrt{n}}$... Do there exist two bounded sequences $a_1, a_2,\ldots$ and $b_1, b_2,\ldots$ such that for each positive integers $n$  and $m>n$ at least one of the two inequalities $|a_m-a_n|>1/\sqrt{n},$ and $|b_m-b_n|>1/\sqrt{n}$ holds?
The original solution used geometric perspective, marking $a_i,b_i$ as points in cartesian plane  etc, are there other non-geometrical methods to solve this problem? Can someone give a hint atleast? The geometrical solution seems very random to me.
 A: Let's define the points $P_n = (a_n,b_n)$ for $n=1,...,+\infty$ on a 2D cartesian plane.
For each point $P_n$, let's draw a circle $\left(P_n,\frac{1}{2\sqrt{n}}\right)$ with center $P_n$ and radius $\frac{1}{2\sqrt{n}}$.
Then all the circles are not intersecting. Indeed, WOLG,  take any circle $\left(P_n,\frac{1}{2\sqrt{n}}\right)$ and $\left(P_m,\frac{1}{2\sqrt{m}}\right)$ with $m>n$ ,  we have
$$P_nP_m=\sqrt{(a_m-a_n)^2+(b_m-b_n)^2}>\sqrt{\left(\frac{1}{\sqrt{n}}\right)^2}=\frac{1}{\sqrt{n}}>\frac{1}{2\sqrt{n}}+\frac{1}{2\sqrt{m}} \tag{1}$$
So, we deduce that all the circles can not be bounded, because their total area is
$$S:=\sum_{n=1}^{+\infty}\pi\left( \frac{1}{2\sqrt{n}}\right)^2=\frac{\pi}{4}\sum_{n=1}^{+\infty}\frac{1}{n}\xrightarrow{n\to+\infty}+\infty \tag{2}$$
However, as the sequences $(a_n)_n$ and $(b_n)_n$ are bounded, the points $P_n$ are also bounded by a defined form with limited area, so the circles, with radius maximum equal to $\frac{1}{2}$ are also bounded.
$$\tag{3}$$
From $(2)$ and $(3)$, we have contracdiction. So, these two bounded sequences cannot exist.
