Tensor ranks and symmetric tensor ranks are the same for points with low symmetric tensor rank

Fix integers n >= 1, d >= 2. Let V be an (n + 1)-dimensional vector space over a field with characteristic zero. Fix a symmetric tensor T is an element of S(d)(V) subset of V(circle times d). Here we prove that the tensor rank of T is equal to its symmetric tensor rank if the latter is at most (d + 1)/2.