prove $\frac{1}{2}\mathbf{\nabla (u \cdot u) = u \times (\nabla \times u ) + (u \cdot \nabla)u}$ using index notation I'm having some trouble using index notation to prove the identity 
$$\frac{1}{2}\mathbf{\nabla (u \cdot u) = u \times (\nabla \times u ) + (u \cdot \nabla)u}$$
The closest I can get is by expanding the first term on the RHS, which gives
$$\mathbf{u \times (\nabla \times u)} = 2u_j \partial x_i u_j - u_j\partial x_i u_i - u_i \partial x_j u_j$$
but I don't see what to do from here (if what I've done so far is correct).
Any help will be appreciated!
EDIT
The comments so far are all a bit dubious about my expression for the first term on the RHS, here's my work:
$$\mathbf{u \times (\nabla \times u)} = \epsilon_{ijk}u_j\epsilon_{klm}\partial x_lu_m$$
now I move the second Levi-Civata symbol to the left and use the identity GFR posted to get 
$$\epsilon_{ijk}u_j\epsilon_{klm}\partial x_lu_m = (\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})u_j\partial x_lu_m$$
expanding this gives
$$\mathbf{u \times (\nabla \times u)} = u_j\partial x_i u_j - u_j\partial x_j u_i$$
This next step i'm not sure about, I move the derivatives to the left of each term
$$u_j\partial x_i u_j - u_j\partial x_j u_i = \partial x_iu_ju_j - \partial x_j u_ju_i$$
then the product rule gives my original equation for $\mathbf{u \times (\nabla \times u)}$
 A: So, your first step is indeed correct, but the one you are not sure about is wrong. You are probably confused by your own notation, I prefer to write $\partial_i=\frac{\partial}{\partial x_i}$ instead of $\partial x_i$. And remember that repeated indices are being summed over, so e.g. $u_ju_j=\mathbf{u}\cdot\mathbf{u}$. Given that, you have
\begin{equation}
u_j\partial _i u_j-u_j\partial_j u_i
=(1/2)\partial_i(u_j u_j)-(\mathbf{u}\cdot \nabla)u_i
=(1/2) \partial_i (\mathbf{u}\cdot\mathbf{u})-(\mathbf{u}\cdot \nabla)u_i,
\end{equation}
which (taking into account your earlier work) is the $i$-th component of the equation
\begin{equation}
\mathbf{u}\times(\nabla\times \mathbf{u})=\frac{1}{2}\nabla(\mathbf{u}\cdot\mathbf{u})-(\mathbf{u}\cdot \nabla)\mathbf{u}
\end{equation}
A: Hint:
Let ${\bf e}_i$ be orthonormal unit vectors then in index notation $${\bf u} = u_i {\bf e}_i$$
The dot product between two vectors can then be written
$${\bf a}\cdot {\bf b}  = a_ib_i$$
in particular when ${\bf a} = {\bf \nabla}$ we get ${\bf \nabla}\cdot {\bf b}  = \partial_ib_i$. The curl can be written
$${\bf a}\times {\bf b} = \epsilon_{kij}a_i b_j {\bf e}_k$$
where $\epsilon_{ijk}$ is the Levi-Civita symbol.
Using this then the first term on the right hand side in your equation becomes
$${\bf u}\times(\nabla\times {\bf u}) = \epsilon_{kij}u_i(\nabla\times {\bf u})_j{\bf e}_k = \epsilon_{kij}\epsilon_{jlm}u_i\partial_lu_m{\bf e}_k$$
Now repeat this for the other terms and compare. You will have to use the identity (the one given by GFR in the comments)
$$\epsilon_{ijk}\epsilon_{imn} = \delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km}$$
to finish it up.
A: It isn't simpler to show as this?
$$
\frac{1}{2}\nabla(\mathbf{u}\cdot\mathbf{u})=\frac{1}{2}\partial_i(u_j^2)=u_j(\partial_iu_j-\partial_ju_i)+u_j\partial_ju_i=\varepsilon_{ijk}[u_j(\partial_iu_j-\partial_ju_i)_k]+(u_j\partial_j)u_i=\mathbf{u}\times(\nabla\times\mathbf{u})+(\mathbf{u}\cdot\nabla)\mathbf{u}.
$$
