How can I prove geometrically: If AM is median of triangle ABC, then: AM < (AB+BC)/2
Draw a triangle with vertices A, B and C. A median of a triangle is a line segment connecting a vertex to the midpoint of the opposing side. As such, each triangle has three medians.
Let's say M is the midpoint on AC. Naturally, $\displaystyle 2AM=AC$. We also know that the sum of the lengths of any two sides of a triangle is (always) strictly greater than the length of the third side (this applies to all triangles!). Thus, $\displaystyle AC=2AM<AB+BC\implies AM<\dfrac{AB+BC}{2}$.