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 $a=3b$.

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 $d(a,c)\le0.675(d(a,b)+d(b,c))$.

$d$ is also a metric. Proof:

First let us call a metric on natural numbers $d$ such that $d(a,b)<1$ and $d(a,b)$ is a rational number for all $a,b$ a "rational metric". Second let $d_1,d_2$ be two rational metrics such that if we set $d=d_1+d_2-d_1 d_2$ then for all $a \neq c, a \neq b$ we have $d(a,b)+d(a,c)>1$. If this is the case for $d_1,d_2$ we will call $d_1$ and $d_2$ "paired". If $d_1,d_2$ are such paired rational metrics, then $d=d_1+d_2-d_1 d_2$ is a metric. Proof:

1) $d(a,b) = 0$ iff $0 \le d_1(a,b)(1-d_2(a,b)) = -d_2(a,b) \le 0$ hence since $1-d_2(a,b)>0$ we must have $d_1(a,b) = 0$ hence $a=b$. If on the other hand $a=b$ then plugging this in $d$ and observing that $d_1(a,b)=d_2(a,b)=0$ gives us $d(a,b)=0$.

2) $d(a,b) = d(b,a)$ since $d_i(a,b) = d_i(b,a)$ for $i = 1,2$.

3) Triangle inequality: If $a=c$ or $a=b$ the triangle inequality is fullfilled and becomes an equality because of 1) : $d(b,c) \le d(a,b)+d(a,c)$ First observe that $d(x,y) < 1$ for all $x,y$. Let therefore $a\neq c, a\neq b$. Since $d_1,d_2$ are paired rational metrics we have: $d(b,c) < 1 < d(a,c)+d(a,b)$ and the triangle inequality is proved.

This proves also that $d$ is a rational metric (if $d_1,d_2$ are paired rational metrics.)

What remains to show is that $d_2(a,b) = 1-\frac{2 \gcd(a,b)}{a+b}$, $d_1(a,b) = 1-\frac{\gcd(a,b)^2}{ab}$ are paired (rational) metrics, hence it remains to show that $d(a,b) = 1- \frac{2 \gcd(a,b)^3}{ab(a+b)}$ satisfies:

$$d(a,c)+d(a,b)>1, \text{ whenever } a\neq c, a \neq b$$

The last inequality is equivalent to, after some algebra:

$$\frac{abc(a+b)(a+c)}{2} - \gcd(a,b)^3c(a+c) - \gcd(a,c)^3b(a+b)>0$$

Let $U=\gcd(a,b,c)$. Then there exist natural numbers $R,S,T,A,B,C$ such that:

$$RU = \gcd(a,b), SU = \gcd(a,c), TU = \gcd(b,c), a = RSUA, b = RTUB, c = STUC$$

Plugging this in the last inequality and after some algebra, we find:

$$1/2*(A^3*B*C*R^2*S^2*T + A^2*B^2*C*R^2*S*T^2 + A^2*B*C^2*R*S^2*T^2 + A*B^2*C^2*R*S*T^3 - 2*A*C*R^2 - 2*A*B*S^2 - 2*C^2*R*T - 2*B^2*S*T)*R^2*S^2*T*U^5 > 0 $$

We can pair each of the positive summand with a negative summand to give for example:

$$(A^3*B*C*R^2*S^2*T-2*A*C*R^2)=(A^2*B*S^2*T - 2)*A*C*R^2$$

The condition $a \neq b$ translates to $SA \neq TB$ and similarily $a \neq c$ translates to $RA \neq TC$. Suppose that $A^2*B*S^2*T - 2 \le 0$. The case $A^2*B*S^2*T=1$ contradicts $SA \neq TB$. Hence we can only have at most $A^2*B*S^2*T=2$ which leads to $A=S=1$, $BT=2$ and plugging this in the definition of $a,b$ we get $b=2a$ and $d(a,b)=\frac{2}{3}$.

Now we must show that the other pairings give the desired result:

$$( A^2*B^2*C*R^2*S*T^2-2*B**2*S*T)=(A^2*C*R^2*T - 2)*B^2*S*T$$ A similar argument to the above leads to: If $A^2*C*R^2*T = 2$, then $A=R=1$, $CT=2$ which leads to (with $S=A=1$) $a=RSUA=U,b=RTUB=2U=2a,c=STUC=2U=2a$ and it follows that $d(a,c)=\frac{2}{3}$, so $d(a,b)+d(a,c)=\frac{4}{3}>1$, and this case is done.

If $A^2*C*R^2*T > 2$ and $A^2*B*S^2*T=2$ then $$1/2*(A^3*B*C*R^2*S^2*T + A^2*B^2*C*R^2*S*T^2 + A^2*B*C^2*R*S^2*T^2 + A*B^2*C^2*R*S*T^3 - 2*A*C*R^2 - 2*A*B*S^2 - 2*C^2*R*T - 2*B^2*S*T)*R^2*S^2*T*U^5 > 0 $$ is true.

If $A^2*C*R^2*T > 2$ and $A^2*B*S^2*T>2$ then $$1/2*(A^3*B*C*R^2*S^2*T + A^2*B^2*C*R^2*S*T^2 + A^2*B*C^2*R*S^2*T^2 + A*B^2*C^2*R*S*T^3 - 2*A*C*R^2 - 2*A*B*S^2 - 2*C^2*R*T - 2*B^2*S*T)*R^2*S^2*T*U^5 > 0 $$ is true. This shows, that $d_1,d_2$ are paired metrics and completes the proof.