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*](https://en.wikipedia.org/wiki/Abc_conjecture#Formulations). 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*](https://math.stackexchange.com/questions/3648776/a-weak-form-of-the-abc-conjecture-involving-the-definition-of-h%c3%b6lder-mean) 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*](https://math.stackexchange.com/questions/3580506/weaker-than-abc-conjecture-invoking-the-inequality-between-the-arithmetic-and-lo) 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* $1\leq a<b$ *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*](https://en.wikipedia.org/wiki/Generalized_mean#Special_cases) and [*Logarithmic mean*](https://en.wikipedia.org/wiki/Logarithmic_mean#Inequalities). I add references to bilbiography. I know the statement of formulation ABC conjecture II for example from [3].


References:
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[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).



On weaker forms of the abc conjecture from the theory of Hölder and logarithmic means
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 $\log(b-a) < (1-\epsilon) \log(b+a)$


