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Oct 10, 2019 at 9:04 comment added Đào Thanh Oai @GerryMyerson Why the answer true? I think the answer is not valid. Because the case $2.6 \le 3.4$ so I can write: $(1−1/2)(1−1/6)=5/12≤1/2=(1−1/3)(1−1/4)$ .
Oct 10, 2019 at 9:00 comment added Gerry Myerson It seems to me that if the conjecture with less-than-or-equal fell to the first possible counterexample, then it's a good idea to check for simple counterexamples before making a conjecture with less-than. There is no evidence that this has been done. Why present a conjecture with no reason to think it's true?
Oct 10, 2019 at 7:38 comment added Đào Thanh Oai @BjørnKjos-Hanssen ecause $2.6 \le 3.4$ we can write $ (1-1/2)(1-1/6)=5/12 \le 1/2=(1-1/3)(1-1/4) $
Oct 10, 2019 at 7:37 comment added Đào Thanh Oai @BjørnKjos-Hanssen Ok, But I think your answer (now) is not answer. Because $a.b=c.d$ $\Rightarrow$ $f(a,b)>f(c,d)$ or $f(a,b)<f(c,d)$ also true.
Oct 10, 2019 at 7:32 comment added Bjørn Kjos-Hanssen I guess we can have one answer for each version, see meta.mathoverflow.net/questions/2951/…
Oct 10, 2019 at 7:31 history edited Bjørn Kjos-Hanssen CC BY-SA 4.0
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Oct 10, 2019 at 7:29 comment added Đào Thanh Oai @BjørnKjos-Hanssen I change my question. If You delete the answer, I will delete version 1 thank you. I am sorry.
Oct 10, 2019 at 6:55 comment added Đào Thanh Oai Dear @BjørnKjos-Hanssen , because $2.6 \ge \le = 3.4$ so $(1-1/2)(1-1/6)=5/12 \not\ge 1/2=(1-1/3)(1-1/4)$ is not counter example.
Oct 10, 2019 at 6:39 comment added Đào Thanh Oai do I want correct $\prod_{i=1}^n a_i \le \prod_{j=1}^k b_j $ to $\prod_{i=1}^n a_i < \prod_{j=1}^k b_j $ and $a_i \le b_j$. @GerryMyerson
Oct 10, 2019 at 5:39 comment added Gerry Myerson In other words, the simplest possible example satisfying the hypothesis already fails to satisfy the conclusion.
Oct 10, 2019 at 5:31 history answered Bjørn Kjos-Hanssen CC BY-SA 4.0