I think I found counter example for this, ie. set of points of which one is inside the 3D convex hull, but it's appropriate contenstant can win. If we are given for contestant with following speeds:

1 1 2
1 1 4
100 1 3
1 100 3
20 20 3

If we choose paths of length 100, 100, 3, than last contestant (20, 20, 3) can win with time 100/20 + 100/20 + 3/3 = 11. But, for this points 3D convex hull consists of (1, 1, 2), (1, 1, 4), (100, 1, 3) and (1, 100, 3).

Does anyone know if I'm wrong and why?

Consider that the sum of swim,bike and run is a const number,so we can let the total s=1,then we can solve it for 2D convex hull alogrithm rather than 3D.

You should use reciprocals of speeds (preferably multiplied by some constant) as coordinates, but not just the speeds. After applying such an "inversion" to the points in your example, the last one jumps out of the convex hull formed by others.