Prove that among any five distinct real numbers there are two, $a$ and $b$, such that $|ab+1|\gt|a-b|$. Solution without trigonometry

Question

Prove that among any five distinct real numbers there are two, $a$ and $b$, such that $\lvert ab+1\lvert
\ \gt \ \vert a-b \vert$.

Solution: I have solved the above problem using Trigonometrical substitution by writing $-90^\circ <\tan\ x_k<90^\circ$, $k=1,2,3,4,5$, and considering the intervals $(-90^\circ,-45^\circ)$; $(-45^\circ,0^\circ)$; $(0^\circ,45^\circ)$; $(45^\circ,90^\circ)$. Then by Pigeonhole Principle , at least two of $x_1,x_2,x_3,x_4,x_5$ will lie in the same interval. Let those two be $(x_i,x_j)$ as $\vert x_i-x_j\vert\lt 45^\circ$ setting $a=\tan x_i$ and $b=\tan x_j$ we get the desired inequality. $\square$

However, I was wondering whether we could approach this problem in an algebraic way because this question is in an Algebraic Book which I have.

Any help would be appreciated.
Thanks!

Answer 1

Essentially the same argument can be made algebraically: at least one of the intervals $(\leftarrow,-1)$, $[-1,0)$, $[0,1)$, and $[1,\to)$ must contain two of the numbers, say $a$ and $b$. We may assume that $a<b$ and let $d=b-a$. (Note that your four intervals are not quite exhaustive, since all of them are open.) Taking the intervals in reverse order:

• if $1\le a<b$, then $$|ab+1|=a^2+ad+1\ge d+2>d=|a-b|\,;$$
• if $0\le a<b<1$, then $$|ab+1|=a^2+ad+1\ge 1>d=|a-b|\,;$$
• if $-1\le a<b<0$, then $$|ab+1|=ab+1>1>d=|a-b|\,;\text{ and}$$
• if $a<b<-1$, then $$|ab+1|=b^2-bd+1>2-bd=2+|b|d>d=|a-b|\,.$$