I hope it is appropriate to ask this question here:


One formulation of the abc-conjecture is 


$$ c < \text{rad}(abc)^2$$


where $\gcd(a,b)=1$ and $c=a+b$. This is equivalent to ($a,b$ being arbitrary natural numbers):


$$ \frac{a+b}{\gcd(a,b)} < \text{rad}(\frac{ab(a+b)}{\gcd(a,b)^3})^2$$


Let $d\_1(a,b) = 1- \frac{\gcd(a,b)^2}{ab}$ which is a proven metric on natural numbers.
Let $d\_2(a,b) = 1- 2 \frac{\gcd(a,b)}{a+b}$, which I *suspect* to be a metric on natural numbers, but I have not proved it yet. Let
$$d(a,b) = d\_1(a,b)+d\_2(a,b)-d\_1(a,b)d\_2(a,b) = 1-2\frac{\gcd(a,b)^3}{ab(a+b)}$$


Then we get the equivalent formulation of the inequality above:


$$\frac{2}{1-d\_2(a,b)} < \text{rad}(\frac{2}{1-d(a,b)})^2$$


which is equivalent to :


$$\frac{2}{1-d\_2(a,b)} < \text{rad}(\frac{1}{1-d\_1(a,b)}\cdot\frac{2}{1-d\_2(a,b)} )^2$$


My question is if one can prove that $d\_2$ and $d$ are distances on the natural numbers (without zero)?


**Result**: By the answer of @GregMartin, $d\_2$ is a metric. By the other answer $d$ is also a metric.


**Edit**:
By "symmetry" in $d\_1$ and $d\_2$, this interpretation also suggests that the following inequality is true , which might be trivial to prove or very difficult or might be wrong and may be of use or not in number theory:


$$\frac{1}{1-d\_1(a,b)} < \text{rad}(\frac{2}{1-d(a,b)})^2$$


which is equivalent to 


$$ \frac{ab}{\gcd(a,b)^2} < \text{rad}(\frac{ab(a+b)}{\gcd(a,b)^3})^2$$


(This is not easy to prove, as the $abc$ conjecture $c=a+b < ab < \text{rad}(abc)^2$ would follow for all $a,b$ such that $a+b < ab$.)


**Second edit**:
Maybe the proof that $d\_2,d$ are distances can be done with some sort of metric transformation, for example maybe with a Schoenberg transform (See 3.1, page 8 in <https://arxiv.org/pdf/1004.0089.pdf>) The idea, that this might be proved with a Schoenberg transform comes from the fact that:


$$d\_1(a,b) = 1-\exp(-\hat{d}(a,b))$$
so $d\_1$ is a Schoenberg transform of $\hat{d}(a,b) = \log( \frac{ab}{\gcd(a,b)^2}) = \log( \frac{\text{lcm}(a,b)}{\gcd(a,b)})$ which is proved to be a metric (see Encyclopedia of Distances, page 198, 10.3 )


**Third edit**:
Here is some Sage Code to test the triangle inequality for triples (a,b,c) up to 100:



```
def d1(a,b):
    return 1-gcd(a,b)**2/(a*b)

def d2(a,b):
    return 1-2*gcd(a,b)/(a+b)

def d(a,b):
    return d1(a,b)+d2(a,b)-d1(a,b)*d2(a,b)

X = range(1,101)
for a in X:
    for b in X:
        for c in X:
            if d2(a,c) > d2(a,b)+d2(b,c):
                print "d2",a,b,c
            if d(a,c) > d(a,b)+d(b,c):
                print "d",a,b,c

```

so far with no counterexample.


Related:
[An inequality inspired by the abc-conjecture and two questions](https://mathoverflow.net/questions/342793/an-inequality-inspired-by-the-abc-conjecture-and-two-questions)



A reinterpretation of the $abc$ - conjecture in terms of metric spaces?
$d\_2$ is indeed a metric. Abbreviating $\gcd(m,n)$ to $(m,n)$, we need to show that
\begin{align\*}
1-\frac{2(a,c)}{a+c} &\le 1-\frac{2(a,b)}{a+b} + 1-\frac{2(b,c)}{b+c}
\end{align\*}
or equivalently
\begin{align\*}
\frac{2(a,b)}{a+b} + \frac{2(b,c)}{b+c} &\le 1 + \frac{2(a,c)}{a+c}.
\end{align\*}
Furthermore, we may assume that $\gcd(a,b,c)=1$, since we can divide everything in sight by that factor.


Note that if $a=(a,b)\alpha$ and $b=(a,b)\beta$ with $(\alpha,\beta)=1$, then $\frac{2(a,b)}{a+b} = \frac2{\alpha+\beta}$. The only unordered pairs $\{\alpha,\beta\}$ for which this is at least $\frac12$ are $\{1,1\}$, $\{1,2\}$, and $\{1,3\}$. Further, if neither $\frac{2(a,b)}{a+b}$ nor $\frac{2(b,c)}{b+c}$ is at least $\frac12$, then the inequality is automatically valid because of the $1$ on the right-hand side.


This leaves only a few cases to check. The case $\{\alpha,\beta\} = \{1,1\}$ (that is, $a=b$) is trivial. The case $\{\alpha,\beta\} = \{1,2\}$ (that is, $b=2a$) can be checked: we have $(a,c)=\gcd(a,2a,c)=1$, and so the inequality in question is
\begin{align\*}
\frac23 + \frac{2(2,c)}{2a+c} &\le 1 + \frac2{a+c},
\end{align\*}
or equivalently
$$
\frac{(2,c)}{2a+c} \le \frac16 + \frac1{a+c};
$$
there are only finitely many ordered pairs $(a,c)$ for which the left-hand side exceeds $\frac16$, and they can be checked by hand.


The proof for the case $\{\alpha,\beta\} = \{1,3\}$ (that is, $b=3a$) can be checked in the same way, as can the cases $a=2b$ and $a=3b$.


