Derivation of the balance of kinetic energy from the momentum balance using the continuity equation 
Derive the balance of kinetic energy from the momentum balance using
the continuity equation.

The momentum equation in 3D reads as follows:
$$ \frac{\partial\rho \mathbf{v}}{\partial t} + \nabla\cdot(\rho \mathbf{v}\otimes\mathbf{v}) = \nabla\cdot\sigma + \mathbf{f_b}. $$
The continuity equation is given by:
$$\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v}) = 0.$$
In the exercise there is a hint to take the inner product of the velocity $\mathbf{v}$ with both
sides of the momentum equation. Then to rewrite that with the help of the continuity equation to obtain the balance equation for
the kinetic energy density:
$$\frac{\partial }{\partial t}\left(\frac{1}{2}\rho v^2\right) = -\nabla\cdot\left(\frac{1}{2}\rho v^2\mathbf{v}\right) + \mathbf{v}\cdot(\nabla\cdot\sigma) + \mathbf{v}\cdot\mathbf{f_b}.$$
Here $v^2 := \mathbf{v}\cdot\mathbf{v}$.
What I've tried is to take the inner product with $\mathbf{v}$ and apply the product rule for derivatives:
\begin{align}
  \mathbf{v}\cdot\left(\rho\frac{\partial\mathbf{v}}{\partial t}\right) + \mathbf{v}\cdot\left(\frac{\partial\rho}{\partial t}\mathbf{v}\right) + \mathbf{v}\cdot(\rho\mathbf{v}\cdot\nabla\mathbf{v}) + \mathbf{v}\cdot(\mathbf{v}\nabla\cdot(\rho\mathbf{v})) = \mathbf{v}\cdot(\nabla\cdot\sigma) + \mathbf{v}\cdot\mathbf{f_b}   
\end{align}
Now the second and fourth term on the LHS should cancel out due to the continuity equation, right? But how will I now ever arrive at the correct equation? I don't get where the halfs are coming from. Should I maybe split the second term into half and substitute $\frac{\partial \rho}{\partial t} = - \nabla\cdot(\rho\mathbf{v})$ into only one of them? I still don't see how some of the other terms cancel out then. Help would be much appreciated.
 A: Taking the dot product of $\mathbf{v}$ and the left-hand side of the momentum equation and using Cartesian components with the Einstein convention for summation over repeated indexes, we get
$$\tag{1}\mathbf{v} \cdot \left[\frac{\partial \rho \mathbf{v}}{\partial t} + \nabla \cdot (\rho \mathbf{v}\mathbf{v}) \right]= v_i \frac{\partial}{\partial t}(\rho v_i)+ v_i\frac{\partial}{\partial x_j}(\rho v_j v_i) \\= v_i^2 \frac{\partial\rho}{\partial t}+ \rho v_i\frac{\partial v_i}{\partial {t}}+ v_i^2\frac{\partial}{\partial x_j}(\rho v_j) + \rho v_j v_i\frac{\partial v_i}{\partial x_j}$$
By the product rule it follows that
$$ \tag{2}\frac{\partial }{\partial t}\left(\frac{1}{2}v_i^2\right)=\frac{1}{2} \frac{\partial v_i^2}{\partial t} = \frac{1}{2}\left[2 v_i\frac{\partial v_i}{\partial t}\right] =  v_i\frac{\partial v_i}{\partial t}, \\ \frac{\partial }{\partial x_j}\left(\frac{1}{2}v_i^2\right)=\frac{1}{2} \frac{\partial v_i^2}{\partial x_j} = \frac{1}{2}\left[2 v_i\frac{\partial v_i}{\partial x_j}\right] =  v_i\frac{\partial v_i}{\partial x_j}$$
Substituting into (1) with (2), we obtain
$$\mathbf{v} \cdot \left[\frac{\partial \rho \mathbf{v}}{\partial t} + \nabla \cdot (\rho \mathbf{v}\mathbf{v}) \right]\\= v_i^2 \frac{\partial\rho}{\partial t}+ \rho \frac{\partial }{\partial {t}}\left(\frac{1}{2}v_i^2\right)+ v_i^2\frac{\partial}{\partial x_j}(\rho v_j) + \rho v_j \frac{\partial }{\partial x_j}\left(\frac{1}{2}v_i^2\right)\\ = v_i^2 \frac{\partial\rho}{\partial t}+ v_i^2\frac{\partial}{\partial x_j}(\rho v_j) + \rho \frac{\partial }{\partial {t}}\left(\frac{1}{2}v_i^2\right)+ \rho v_j \frac{\partial }{\partial x_j}\left(\frac{1}{2}v_i^2\right)\\ = \frac{1}{2} \left[v_i^2 \frac{\partial\rho}{\partial t}+ v_i^2\frac{\partial}{\partial x_j}(\rho v_j)  \right]+  \frac{1}{2} \left[v_i^2 \frac{\partial\rho}{\partial t}+ v_i^2\frac{\partial}{\partial x_j}(\rho v_j)  \right]+  \rho \frac{\partial }{\partial {t}}\left(\frac{1}{2}v_i^2\right)+ \rho v_j \frac{\partial }{\partial x_j}\left(\frac{1}{2}v_i^2\right)\\ = \frac{v_i^2}{2} \left[ \frac{\partial\rho}{\partial t}+ \frac{\partial}{\partial x_j}(\rho v_j)  \right]+  \frac{1}{2} v_i^2 \frac{\partial\rho}{\partial t}+  \rho \frac{\partial }{\partial {t}}\left(\frac{1}{2}v_i^2\right)+  \frac{1}{2}v_i^2\frac{\partial}{\partial x_j}(\rho v_j) +\rho v_j \frac{\partial }{\partial x_j}\left(\frac{1}{2}v_i^2\right)$$
By the continuity equation, the first term on the RHS vanishes and we get
$$\tag{3}\mathbf{v} \cdot \left[\frac{\partial \rho \mathbf{v}}{\partial t} + \nabla \cdot (\rho \mathbf{v}\mathbf{v}) \right]\\=\frac{1}{2} v_i^2 \frac{\partial\rho}{\partial t}+  \rho \frac{\partial }{\partial {t}}\left(\frac{1}{2}v_i^2\right)+  \frac{1}{2}v_i^2\frac{\partial}{\partial x_j}(\rho v_j) +\rho v_j \frac{\partial }{\partial x_j}\left(\frac{1}{2}v_i^2\right),$$
By the product rule we can combine the second and third terms on the RHS of (3) as
$$\frac{1}{2} v_i^2 \frac{\partial\rho}{\partial t}+  \rho \frac{\partial }{\partial {t}}\left(\frac{1}{2}v_i^2\right) = \frac{\partial}{\partial t}\left(\frac{1}{2}\rho v_i^2\right), $$
and similarly we can combine the fourth and fifth terms on the RHS of (3) to get
$$\frac{1}{2}v_i^2\frac{\partial}{\partial x_j}(\rho v_j) +\rho v_j \frac{\partial }{\partial x_j}\left(\frac{1}{2}v_i^2\right) = \frac{\partial}{\partial x_j} \left[ \left(\frac{1}{2} \rho v_i^2\right) v_j \right]$$
Substituting into (3) yields
$$\mathbf{v} \cdot \left[\frac{\partial \rho \mathbf{v}}{\partial t} + \nabla \cdot (\rho \mathbf{v}\mathbf{v}) \right]\\=\frac{\partial}{\partial t}\left(\frac{1}{2}\rho v_i^2\right)+\frac{\partial}{\partial x_j} \left[ \left(\frac{1}{2} \rho v_i^2\right) v_j \right] \\ = \frac{\partial}{\partial t}\left(\frac{1}{2}\rho |\mathbf{v}|^2\right)+\nabla \cdot \left(\frac{1}{2} \rho |\mathbf{v}|^2 \mathbf{v} \right),$$
where under the Einstein convention $v_i^2 = v_1^2+v_2^2 + v_3^2 = |\mathbf{v}|^2$ (the same as $v^2$ in your notation).
