Poisson brackets of angular momentum [closed]

The Poisson bracket of two functions $f$ and $g$ is defined as $$\{f,g\}=\sum\limits_l\left(\frac{\partial f}{\partial x_l}\frac{\partial g}{\partial p_l}-\frac{\partial f}{\partial p_l}\frac{\partial g}{\partial x_l}\right).$$ I want to show that $$\{L_i,F(\vec{x})\} = [\vec{x}\times\vec{\nabla}]_i F(\vec{x}),$$ where $L_i$ is a component of the angular momentum vector $\vec{L} = \vec{x}\times\vec{p}$ and $F(\vec{x})$ is some arbitrary function. Using that $$L_i = \epsilon_{ijk}x_jp_k,$$ I arrive at (note that $\partial_i = \displaystyle\frac{\partial}{\partial x_i}$) \begin{align}\{L_i,F(\vec{x})\} &= \sum\limits_l\left(\frac{\partial (\epsilon_{ijk}x_jp_k)}{\partial x_l}\frac{\partial F(\vec{x})}{\partial p_l}-\frac{\partial (\epsilon_{ijk}x_jp_k)}{\partial p_l}\frac{\partial F(\vec{x})}{\partial x_l}\right) \\ &= -\epsilon_{ijk}x_j\partial_k F(\vec{x}) \\& = -[\vec{x}\times\vec{\nabla}]_i F(\vec{x})\end{align} (since the first term vanished because $F(\vec{x})$ does not depend on $\vec{p}$) which is almost the desired result. The exercise asks me explicitly to prove the formula without the minus sign. So, what could be the reason why I get one and how could I get rid of that problem?

• Why do you think you can get "rid" of the minus sign? Your result is correct. Dec 6, 2016 at 15:39
• @ACuriousMind The trouble is that the formula I have to prove differs from the formula I derived by a minus sign. I've edited my post to make that clearer (hopefully). Dec 6, 2016 at 16:05