All of the variables are vectors. It is based on the lagrangian of a electromagnetic field, where v is the velocity and A the cross potential.
I've tried to prove it by using their modulus, sines and cosines but I don't think that's the way to prove it.
Could you give me any hint?
Thanks in advance.
Let's use the summation convention. Let ${\bf v}= v_i{\bf e}_i$, ${\bf A}=A_i{\bf e}_i$. We have, on the right side $$ \nabla \times {\bf A} = {\bf e}_i\epsilon_{ijk}\partial_jA_k$$ $$ {\bf v}\times (\nabla \times {\bf A}) = {\bf e}_i\epsilon_{ijk}v_j(\nabla \times {\bf A})_k={\bf e}_i\epsilon_{ijk}v_j\epsilon_{kmn}\partial_mA_n $$ Using the property $\epsilon_{ijk}\epsilon_{kmn}=\delta_{im}\delta_{jn}-\delta_{in}\delta_{jm}$ we get $$ {\bf v}\times (\nabla \times {\bf A}) = (\delta_{im}\delta_{jn}-\delta_{in}\delta_{jm}){\bf e}_iv_j\partial_mA_n = {\bf e}_i(v_j\partial_iA_j - v_j\partial_jA_i )$$ It is generally NOT equal the left side, which is $$ \nabla({\bf v}\cdot{\bf A}) - ({\bf v}\cdot\nabla){\bf A} = {\bf e}_i\partial_i (v_jA_j) - (v_j\partial_j)({\bf e}_iA_i) = {\bf e}_i \big(\partial_i (v_jA_j) - v_j\partial_jA_i\big)$$ As we can see, the equality occurs only when $$ v_j\partial_iA_j = \partial_i (v_jA_j)$$ that is when $$ (\partial_i v_j)A_j = 0$$