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 $$|ab+1|=b^2-bd+1>2-bd=2+|b|d>d=|a-b|\,.$$

Answer 2

Your use of the tangent function is a good inspiration from your knowledge of trigonometry, but you can avoid it just by making the intervals be for the $x_k$ instead of their tangents and making the intervals $(-\infty,-1], (-1,0],(0,1],(1,\infty)$. Now show that if two of the points are in the same interval the desired inequality is satisfied. For example, if $a,b \in (0,1]$ we have $$|ab+1|=ab+1\gt 1 \gt |a-b|$$

You can prove it for the other three intervals as well. Someone who does not know any trig might think you had a miraculous inspiration to choose these intervals, but $0$ and $\pm 1$ are often special when multiplication is involved.

Answer 3

You don't have to speak of $5$ distinct real numbers, since you're not using the remaining three But instead I suggest, $2$ distinct number $$\lvert ab+1\lvert \ \gt \ \vert a-b \vert$$ $$ (ab+1)^2 \gt (a-b)^2 $$ $$ a^2b^2+2ab+1 \gt a^2-2ab+b^2$$ $$ a^2b^2+2ab+1+4ab \gt a^2-2ab+b^2+4ab$$ $$a^2b^2+6ab+1 \gt a^2+2ab+b^2$$ $$a^2b^2+6ab+1 \gt ( a+b)^2 $$ $$ab( ab+6)+1 \gt (a+b)^2 $$ It's easy to know that $ ab \gt a+b$, for $a,b \in \mathbb{Z}$ $$ab \cdot (ab+6) +1 \gt (a+b) \cdot (a+b) $$ Then the identity is true, or are you trying to proof it with trigonometry alone