About $\nabla$'s Property. For a scalar function $g$, and a vector function $f$,
$$ | \nabla ( (\nabla g) \cdot f ) | \leqslant |f| \cdot \text{Something} $$
Is this inequality possible? If possible, what would "$\text{Something}$" be?
 A: I dont know exactly what kind of estimative you are looking for. Here it goes one: I assume $f\in\mathbb{R}^3$ is a constant vector, and change the notation $f$ by $v$. We see that
\begin{eqnarray}
\nabla((\nabla g)\cdot v)&=&\nabla\frac{\partial g}{\partial v}\\
&=&\left(\frac{\partial}{\partial x}\frac{\partial g}{\partial v},\frac{\partial}{\partial y}\frac{\partial g}{\partial v},\frac{\partial}{\partial z}\frac{\partial g}{\partial v}\right)\\
&=&\left(\frac{\partial}{\partial v}\frac{\partial g}{\partial x},\frac{\partial}{\partial v}\frac{\partial g}{\partial y},\frac{\partial}{\partial v}\frac{\partial g}{\partial z}\right)\\
&=&\frac{\partial}{\partial v}(\nabla g)\\
&=&\textrm{D}^2g \cdot v
\end{eqnarray}
and hence,
$$|\nabla\left((\nabla g)\cdot v\right)|\leqslant C\cdot|v|$$
where $C=\sup_{w\in\mathbb{S}^2}|\textrm{D}^2g \cdot w|$, or, in other words, $C$ is the spectral radius of $\textrm{D}^2g$, wich of course depends on the point where you are evaluating the inequality.
A: Here it goes an improvement of the last estimative i did before.
Assuming $F(x)=(f_1(x),f_2(x),f_3(x))$ a differentiable vector function of the variable $x=(x_1,x_2,x_3)$, lets calculate $\nabla ( (\nabla g) \cdot F )$. I will use the notation $\frac{\partial}{\partial x_i}g=g_{x_i}$. All the sums will be with $i=1,2,3$:
\begin{eqnarray}
\nabla((\nabla g)\cdot F) &=& \nabla\left(\sum g_{x_i}f_i\right)\\
&=&\left(\left(\sum g_{x_i}f_i\right)_{x_1},\left(\sum g_{x_i}f_i\right)_{x_2},\left(\sum g_{x_i}f_i\right)_{x_3}\right)\\
&=&\sum\left(\begin{array}{c}
g_{x_ix_1}f_i \\
g_{x_ix_2}f_i \\
g_{x_ix_3}f_i \end{array}\right)+\sum\left(\begin{array}{ccc}
g_{x_1}(f_i)_{x_1} \\
g_{x_2}(f_i)_{x_2} \\
g_{x_3}(f_i)_{x_3} \end{array}\right)\\
&=&\textrm{D}^2g\cdot F+A
\end{eqnarray}
where $A$ is the second term beside $\textrm{D}^2g\cdot f$. Lets take a more carefull look at $A$.
\begin{eqnarray}
A&=&\left(\begin{array}{c}
g_{x_1}[(f_1)_{x_1}+(f_2)_{x_1}+(f_3)_{x_1}] \\
g_{x_2}[(f_1)_{x_2}+(f_2)_{x_2}+(f_3)_{x_2}] \\
g_{x_3}[(f_1)_{x_3}+(f_2)_{x_3}+(f_3)_{x_3}] \end{array}\right)\\
&=&\left(\begin{array}{ccc}
A_1 & 0 & 0 \\
0 & A_2 & 0 \\
0 & 0 & A_3 \end{array}\right)\cdot\nabla g
\end{eqnarray}
where
$$A_i=F_{x_i}\cdot(1,1,1)=\sqrt[3]{2}\cdot\left(F_{x_i}\cdot\frac{1}{\sqrt[3]{2}}(1,1,1)\right)=:\sqrt[3]{2}\cdot\left(F_{x_i}\cdot u\right)$$
where $u\in\mathbb{S}^2$ is a unitary vector. Hence,
$$|\nabla((\nabla g)\cdot F)|\leqslant\sup_{w\in\mathbb{S}^2}|\textrm{D}^2g \cdot w|\cdot|F|+|\nabla g|\cdot\max|A_i|$$
It is all indicating we need hypotesys on $g$ and its derivatives, as well as on $F$ and its derivatives (because $A_i$ depends on it) to get some  useful inequality.
