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
Best Answer
$d_2$ 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 $a=3b$.