A reinterpretation of the $abc$ - conjecture in terms of metric spaces? 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
 A: Not an answer but an observation.
Set $r_2(a,b,c)=d_2(a,c)/(d_2(a,b)+d_2(b,c))$ (when defined), and similarly for $r(a,b,c)$.
Then Greg Martin's proof shows that the values of $r_2$ should be discrete, and
indeed experimentally the values are in decreasing order
$(1,9/10,6/7,5/6,9/11,...)$
The same experiment done for $d$ gives
$(1,27/40,40/63,28/45,...)$
Thus, apart from trivial cases such as $a=b$ one should have the stronger triangle inequality
$d(a,c)\le0.675(d(a,b)+d(b,c))$.
A: 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.
A: $d_2$ 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 $a=3b$.
