It can be zero.

Take the standard metric on $\mathbb R^3$ and the one given in [this example][1] by S. Ivanov.
Below I give simplification of his example which works in your case.

**Simplified example.**
Choose two points $x$ and $y$ 
in $\mathbb R^2$ and construct a sequence of arcs $\gamma_n$ between them.
For each $n$ choose small $\varepsilon_n>0$, so that $\varepsilon_n\to 0$ very fast.
Choose disjoint $\varepsilon_n$-intervals on $\gamma_n$, so that it cover all $\gamma_n$ except set of lenght $\varepsilon_n$.
Construct metric $d_1$ so that it is very cheap to go along each such interval, but compensate it by making very expancive to get to such interval, so you can not use it as a shortcut.

Let $d_2$ be the Euclidean metric. 
Then the $d$-length of $\gamma_n$ has order of $\varepsilon_n$. In particular $d(x,y)=0$. 


  [1]: http://mathoverflow.net/questions/125283/on-lipschitz-embeddability-of-certain-compact-metric-spaces-into-mathbbrn/125295#125295