**Proof 2.1**
To prove this, we construct a tree with four leaves, where two
edges are very long and the other edges are very short (see Figure

4 on top). We choose the length of the middle
edge

*e* to be close to zero. We also choose the distances so that

*d*_{AB} +

*d*_{CD} =

*d*_{AC} +

*d*_{BD}. For simplicity we look at

of the distances and scores. As you can see in
Figure

4, both sums of distances are the same:
(

*d*_{AB} +

*d*_{CD})/2 = 70, and
(

*d*_{AC} +

*d*_{BD})/2 = 70.
The scores can be read from the graph. The score
(

*d*_{AC} +

*d*_{BD})/2 is the midpoint of the line between

*S*_{AC} and

*S*_{BD}. The intersection with the y axis gives us the score,
which is about 6.3. We do the same for
(

*S*_{AB} +

*S*_{CD})/2. This
case is simpler, as both values are the same, so we already have
the midpoint. The score is around 4.5. If the graph is not a
straight line, there will be points where the sum of the distances
are the same, and sum of the scores are different.
Now we change the length of edge

*e* slightly by adding a very
small amount.
(

*d*_{AC} +

*d*_{BD})/2 is now slightly greater than
70, but the score
(

*S*_{AC} +

*S*_{BD})/2 is still clearly larger
than the score
(

*S*_{AB} +

*S*_{CD})/2. Even though

*d*_{AB} +

*d*_{CD} <

*d*_{AC} +

*d*_{BD}, the condition for the scores,

*S*_{AB} +

*S*_{CD}
>

*S*_{AC} +

*S*_{BD}, does not hold. If the curvature was negative,
then we would move
(

*d*_{AC} +

*d*_{BD})/2 to be slightly lower than
70. Unless there is no curvature, i.e.

*S*(

*d*) is a straight
line, we can always find a counter example.