Trivial (?) Proof

Question

Take, $a$, $b$, $c$, $d$ to be real numbers. If $a>b$ and $c>b$ and $d>b$, can we prove that TECHNICALLY, $a+c>d$, or $a+d>c$, or $c+d>a$. Kind of a brain teaser but I am looking for some honest proof. Thanks.

Technically, the assumption has nothing to do with the conclusion, and technically it is always satisfied for some $b$. — Arnaud Mortier, Feb 15 '18 at 18:04

score 3 · Answer 1 · answered Feb 15 '18 at 17:51

3

The claim is false: For example, take

$$a=c=d=-1$$ $$b=-10000$$

answered Feb 15 '18 at 17:51

idok

1,701

Perhaps limit the domain to natural numbers? – Gil Keidar Feb 15 '18 at 17:52
@Gil Then it is trivial: Add the largest two. – Andrés E. Caicedo Feb 15 '18 at 17:55
And the OP explicitly wrote "real numbers" ...+1 – DonAntonio Feb 15 '18 at 17:57
@AndrésE.Caicedo Ah, I see your point. – Gil Keidar Feb 15 '18 at 17:59
@DonAntonio I know, I thought that if the domain were restricted then perhaps the conjecture could be proved. – Gil Keidar Feb 15 '18 at 18:00

Trivial (?) Proof

1 Answers1