Okay, you've got a triangle with vertices at u, v, and w. You want to prove it is shorter to go from u straight to w than it would be to go from u to v AND from u to w. Is this a typo? I'll assume you meant: prove UW ≤ UV + VW.
But first, if you did mean what you typed then the proof is simple: G is a space so none of its side lengths can be zero. Because distances are positive values, adding a positive to a number means the sum will be greater than the number you started with.
On the other hand, this problem is essentially proving the shortest distance between two points is a straight line. To help you follow this proof, please draw an obtuse triangle (it doesn't have to be obtuse, but it is easier to see it this way). Label the vertex of the obtuse angle as U and the other vertices as V and W. Now, starting at W, draw a line through U and end it at a point (call it X) such that UX = UV. Mark these as congruent.
Now, UV = UX and UX + UW = XW. This also means UV + UW = XW. Also, triangleUVX is isosceles and angleUVX is equal to angleUXV . Note that angleXVW > angleXVU (because you have to add a positive angle to angleUVX to get angleXVW) and therefore angleXVW > angleUXV There is a threorem state states when an angle in a triangle gets larger, the side opposite it must get longer. Therefore, because angleXVW > angleUXU, XW > VW. This is the same thing as saying UV + UW > VW.
Just in case you didn't follow that, there are a number of videos of this idea on YouTube. Here's one:
Answers & Comments
Verified answer
You typed d(u, w) <= d(u, v) + d(u, w), when I expected d(u, w) <= d(u, v) + d(v, w). As for what you typed, the inequality is very simple, since
0 <= d(u, v) :: add d(u, w) to both sides
d(u, w) <= d(u, v) + d(u, w)
For the version I was expecting, you can essentially square both sides and it will fall out pretty quickly.
Edit: Ah, this is graph theory and not metric space theory. Oops. Since you've asked the same question again I won't edit this answer further.
Okay, you've got a triangle with vertices at u, v, and w. You want to prove it is shorter to go from u straight to w than it would be to go from u to v AND from u to w. Is this a typo? I'll assume you meant: prove UW ≤ UV + VW.
But first, if you did mean what you typed then the proof is simple: G is a space so none of its side lengths can be zero. Because distances are positive values, adding a positive to a number means the sum will be greater than the number you started with.
On the other hand, this problem is essentially proving the shortest distance between two points is a straight line. To help you follow this proof, please draw an obtuse triangle (it doesn't have to be obtuse, but it is easier to see it this way). Label the vertex of the obtuse angle as U and the other vertices as V and W. Now, starting at W, draw a line through U and end it at a point (call it X) such that UX = UV. Mark these as congruent.
Now, UV = UX and UX + UW = XW. This also means UV + UW = XW. Also, triangleUVX is isosceles and angleUVX is equal to angleUXV . Note that angleXVW > angleXVU (because you have to add a positive angle to angleUVX to get angleXVW) and therefore angleXVW > angleUXV There is a threorem state states when an angle in a triangle gets larger, the side opposite it must get longer. Therefore, because angleXVW > angleUXU, XW > VW. This is the same thing as saying UV + UW > VW.
Just in case you didn't follow that, there are a number of videos of this idea on YouTube. Here's one:
http://www.youtube.com/watch?v=0MT-79u0qgM