Maximizing $f$ in $\mathbb{R}^3$ 
Find the domain and the maximum value that the function
  $$f(x,y,z)=\frac{x+2y+3z}{\sqrt{x^2+y^2+z^2}}$$
  may attain in its domain.

I have found the domain of the function to be $\mathbb{R^3\backslash\mathbf{0}}$. To maximize I differentiated in terms of $x,y,z$ having
$$f_x=\frac{-2 x y-3 x z+y^2+z^2}{\left(x^2+y^2+z^2\right)^{3/2}},\quad f_y=\frac{2 x^2-x y+z (2 z-3 y)}{\left(x^2+y^2+z^2\right)^{3/2}},\quad f_z=\frac{3 \left(x^2+y^2\right)-z (x+2 y)}{\left(x^2+y^2+z^2\right)^{3/2}}$$
But to solve the system $f_x=0,f_y=0$ and $f_z=0$ is rather hard. What are the plausible values of $x,y,z$?

By C-S $$(x^2+y^2+z^2)(1^2+2^2+3^2)\geq(x+2y+3z)^2,$$
which gives
$$-\sqrt{14}\leq\frac{x+2y+3z}{\sqrt{x^2+y^2+z^2}}\leq\sqrt{14}.$$
The equality occurs for $(x,y,z)||(1,2,3),$ which gives that we got a maximal value and the minimal value.
