TITLE: A reinterpretation of the $abc$ - conjecture in terms of metric spaces?
QUESTION [23 upvotes]: 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

REPLY [3 votes]: This question has already very good answers. I justed wanted to highlight that it is possible to shorten the proofs, using the following:
If $X_a = \{ a/k | 1 \le k \le a \}$ then $X_a \cap X_b = \gcd(a,b)$, which is straightforward to prove.
Then $d_1(a,b) = 1-\gcd(a,b)^2/(ab) = 1-|X_a \cap X_b|^2 / (|X_a||X_b|)$ is the squared cosine metric (see Encyclopedia of Distances) and $d_2(a,b) = 1-2\gcd(a,b)/(a+b) = 1-2|X_a \cap X_b| / (|X_a|+|X_b|)$ is the Sorensen Metric (Encyclopedia of Distances). Hence $d_1,d_2$ are metrics of the form $d_i = 1- s_i$ where $s_i$ is a similarity.
But then $s=s_1 \cdot s_2$ is also a similarity and $d=d_1 +d_2 -d_1 d_2 = 1-s=1-s_1 s_2$ is a metric.