Call the triangle ABC and M the midpoint of BC so that AM is the median from A to BC.
We are to prove AM<(AB+AC)/2.
Construct point D such that ABDC is a parallelogram.
So we have BD=AC, CD=AB, and AM=MD.
The diagonals of a parallelogram bisect each other so A, M, and D are on the same line and AD=2*AM
Looking at triangle ABD we see AB+BD>AD by the triangle inequality.
Substituting we get
AB+AC>2AM.
or
AM<(AB+AC)/2.

q.e.d.

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