Need to prove that
$$(\vec{v}\cdot\nabla)\vec{v}=\frac 12\nabla(\vec{v} \cdot \vec v) + (\nabla\times\vec v)\times \vec v$$
I could do it by applying the definitions directly, but triple product gives almost the right answer:
$$(\vec a\times \vec b)\times \vec c=-(\vec c\cdot\vec b)\vec a+(\vec c\cdot\vec a)b$$
In my case I get
$$(\vec{v}\cdot\nabla)\vec{v}=\nabla(\vec{v} \cdot \vec v) + (\nabla\times\vec v)\times \vec v$$
But $\frac 12$ is missing from the first member from the right side and I just can't see where that should come from.
Using implied summation notation, the right-hand side maybe written $$\begin{align} \frac12 \nabla(\vec v\cdot \vec v )+(\nabla \times \vec v)\times \vec v & = \frac12 \hat x_i \partial_i (v_jv_j )-(\hat x_k v_k) \times \left((\hat x_l\partial_l) \times (\hat x_mv_m ) \right)\\ & = \frac12 \hat x_i \partial_i (v_jv_j )-[\hat x_k \times (\hat x_l \times \hat x_m )] v_k \partial_l (v_m ) \\ & = \frac12 \hat x_i \partial_i (v_jv_j )-[\delta_{km} \hat x_l-\delta_{kl} \hat x_m] v_k \partial_l (v_m ) \\ & = \frac12 \hat x_i \partial_i (v_jv_j )-[ \hat x_l v_k \partial_l (v_k )- \hat x_m v_k \partial_k (v_m )] \\ & = \frac12 \hat x_i \partial_i (v_jv_j )-[ \frac12 \hat x_l \partial_l (v_kv_k )- v_k \partial_k (\hat x_mv_m )] \\ & = v_k \partial_k (\hat x_mv_m ) \\ & = (\vec v \cdot \nabla)\vec v \end{align}$$
which proves the identity!