# On weaker forms of the abc conjecture from the theory of Hölder and logarithmic means

In this post (the content of this post is now cross-posted from Mathematics Stack Exchange see below) we denote the radical of an integer $$n>1$$ as the product of disctinct primes dividing it $$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\p\text{ prime}}}p,$$ with the definition $$\operatorname{rad}(1)=1$$. The abc conjecture is an important problem in mathematics as one can to see from the Wikipedia abc conjecture. In this post I mean the formulation ABC conjecture II stated in previous link.

I was inspired in the theory of generalized mean or Hölder mean (see [1]) to state the following claim (Mathematics Stack Exchange 3648776 with title A weak form of the abc conjecture involving the definition of Hölder mean asked Apr 28 '20).

Claim. On assumption of the abc conjecture $$\forall \varepsilon>0$$ there exists a constant $$\mu(\epsilon)>0$$ such that for triples of positive integers $$a,b,c\geq 1$$ satisfying $$\gcd(a,b)=\gcd(a,c)=\gcd(b,c)=1$$ and $$a+b=c$$ ones has for real numbers $$q>0$$ that the following inequality holds $$c<\mu(\varepsilon)\left(\frac{\operatorname{rad}(a)^q+\operatorname{rad}(b)^q+\operatorname{rad}(c)^q}{3}\right)^{3(1+\varepsilon)/q}.\tag{1}$$

Remark 1. Thus as $$q\to 0$$ from the theory of Hölder mean we recover the abc conjecture.

In a similar way I was inspired in the definition of the logarithmic mean and its relationship to the arithmetic mean to pose the following conjecture (Mathematics Stack Exchange 3580506 with title Weaker than abc conjecture invoking the inequality between the arithmetic and logarithmic means asked Mar 14 '20).

Conjecture. For every real number $$\varepsilon>0$$, there exists a positive constant $$\mu(\varepsilon)$$ such that for all pairs $$(a,b)$$ of coprime positive integers $$1\leq a the following inequality holds $$2\,\frac{b-a}{\log\left(\frac{b}{a}\right)}\leq \mu(\varepsilon)\operatorname{rad}(ab(a+b))^{1+\varepsilon}.\tag{2}$$

Remark 2. Thus I think that previous conjecture is weaker than the abc conjecture by virtue of the relation between the artihmetic and logarithmic means.

Question. I wondered what work can be done to prove/discuss unconditionally (I mean on assumption of the cited requirements/conditions, but without invoking any formulations of the abc conjecture) the veracity of previous Claim for the smallest $$q>0$$ close to* $$0$$ that you are able to prove. Similarly**, is it possible to prove Conjecture? Many thanks.

*I'm curious to know what is the smallest $$q>0$$ close to $$0$$ such that the inequality in Claim is true, I think that the right discussion is for $$q>0$$ but if you want to discuss $$|q|$$ very close to $$0$$ because you think that it makes sense, feel free to study our inequality for real numbers $$|q|$$ very close to $$0$$.

$$^{**}$$On the other hand I think that should be possible to prove the Conjecture, since I think that this statement is much weaker than the abc conjecture.

I was inspired in the Wikipedia articles for Generalized mean and Logarithmic mean. I add references to bilbiography. I know the statement of formulation ABC conjecture II for example from [3].

## References:

[1] P. S. Bullen, Handbook of Means and Their Inequalities, Dordrecht, Netherlands: Kluwer (2003).

[2] B. C. Carlson, Some inequalities for hypergeometric functions, Proc. Amer. Math. Soc., 17: in page 36 (1966).

[3] Andrew Granville and Thomas J. Tucker, It’s As Easy As abc, Notices of the AMS, Volume 49, Number 10 (November 2002).

• (2/2) I have hope that my inequalities are interesting, feel free to refer these if you know that some colleague (a professor) have studied the abc conjecture. On the other hand I am trying to read other questions posted on this MathOverflow about weak forms of the abc conjecture, and I know that in the literature there are also articles that were written by professors about relaxations of the abc conjecture. May 8, 2020 at 10:21
• abc implies $b-a$ can't be much smaller that $a+b$ and we have $\log(b-a) \sim \log(a+b)$
– joro
May 8, 2020 at 11:23
• Many thanks for your contribution, and many thanks for the upvoter. As aside/unrelated comment the logarithmic mean, as refers the Wikipedia Stolarsky mean, is a particular case of the Stolarsky mean, I say it if can be inspiring for some user. If I understand well Lehmer was a number theorist and Stolarsky studies in particular diophantine approximation. Hölder provided us, in particular, his important inequality as refers the Wikipedia Hölder's inequality (I refer these facts for all users, since I like to study certain mathematical details when I try to evoke certain relationships). May 8, 2020 at 12:53
• Many thanks to the upvoter and for all those persons that read the post for his/her attention. May 29, 2020 at 16:56

abc implies your conjecture with $$b-a$$.

Case 1 Let $$a,b,c=a+b$$ be bad abc triple,i.e. $$c < rad(ab(a+b))$$.

We have $$rad(ab(a+b)) > c > b - a$$.

Case 2 Let $$a,b,c=a+b$$ be good abc triple,i.e. $$c>rad(ab(a+b))$$.

Then $$T : (b-a)^2,4ab,(a+b)^2$$ is good abc triple too.

The radical is divisor of $$ab(a+b)(b-a)$$ and we have $$(a+b)^2 > (a+b)(b-a)$$.

If $$\log(b-a) < (1-C) \log(b+a)$$ this will give infinitely many good abc triples with quality $$2/(2-C)$$, which contradicts abc.

In summary, abc implies there are only finitely many good abc triples satisfying $$\log(b-a) < (1-\epsilon) \log(b+a)$$

• Many thanks for your excellent and interesting answer. I was disconeccted (yesterday) and I don't know if I loss the opportunity to award your answer with the bounty that was expiring/ending (now I don't know if my words make sense, since I don't know how works a bounty for the last day in which is offering: this bounty expires just 7 hours ago and you answer is edited 15 hours ago). May 31, 2020 at 6:39
• I've flagged my (question) post asking, if possible, to award your answer with the bounty, I was disconnected (yesterday) and I don't know if I loss the opportunity to award the available answer while the bounty was expiring (I don't know if now is possible). Isn't required a response and good day. May 31, 2020 at 7:18
• The moderators told me about my flag that they don't have a method to restore bounties. Thus I'm sorry if as a consequence of that I was disconnected I can not award the bounty. Jun 6, 2020 at 14:24