Vector Identities Proof

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Let $\textbf{r}=(x,y,z)$ and $r=||\textbf{r}||$

Show that $\nabla \cdot (r\space\textbf{r})=4r.$

So, I know that we can use a Vector property to expand the LHS so we get: $$(\nabla r)\cdot\textbf{r}+r(\nabla\cdot\textbf{r})$$ and that $r=\sqrt{3}$
But I'm not too sure where to go from here, and to how to calculate $\nabla \cdot \sqrt{3}$
Any help would be great thanks!

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We need two facts: $$\nabla r=(\frac{\partial r}{\partial x},\frac{\partial r}{\partial y},\frac{\partial r}{\partial z})=\mathbf{r}/r$$ since $\frac{\partial}{\partial x}\sqrt{x^2+y^2+z^2}=\frac{x}{\sqrt{x^2+y^2+z^2}}$, etc. $$\nabla\cdot\mathbf{r}=\frac{\partial x}{x}+\frac{\partial y}{y}+\frac{\partial z}{z}=3$$

Then combining together $$\nabla\cdot(r\mathbf{r})=(\nabla r)\cdot\mathbf{r}+r(\nabla\cdot\mathbf{r})=\frac{\mathbf{r}\cdot\mathbf{r}}{r}+3r=4r$$