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

one formulation. $\endgroup$ – user6671 Sep 30 '19 at 4:10YesorNo? $\endgroup$ – Wlod AA Sep 30 '19 at 4:171more comment