The answer is 3/4.
proof
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Let us consider triangle ABC, with centroid G. Let ma, mb and mc be the medians from A, B and C respectively. Then, for instance, AG = 2/3.ma, since G divides the medians in ratio 2:1.
We can apply the triangle inequality to triangles ABG, AGB and GBC, to find:
2/3.ma + 2/3.mb > c (= AB)
2/3.mb + 2/3.mc > a (= BC)
2/3.mc + 2/3.ma > b (= AC)
Adding these three inequalities we find:
4/3.(ma + mb + mc) > a + b + c
ma + mb + mc > 3/4.(a + b + c)