Show that $\|(x,y,z)\| = |x| + 2 \sqrt{y^{2} + z^{2}}$ is a norm on $\mathbb{R}^3$

Question

Show that $\|(x,y,z)\| = |x| + 2 \sqrt{y^{2} + z^{2}}$ is a norm on $\mathbb{R}^3$

My issue comes with showing the 3rd condition, i.e the triangle inequality. After some expansion I am left with:

$$\|(x_1,y_1,z_1) + (x_2,y_2,z_2)\| = |x_{1} + x_{2}| + 2\sqrt{(y_{1}^{2} + z_{1}^{2}) + 2y_{1}y_{2} + 2z_{1}z_{2} + (y_{2}^{2} + z_{2}^{2})}$$

So that middle set of $2y_{1}y_{2} + 2z_{1}z_{2}$ is being a nuisance. Suggestions on what inequality I am not taking advantage of?

Answer 1

By the triangle inequality and by Minkowski we obtain: $$\left\|(x_1,y_1,z_1)\right\|+\left\|(x_2,y_2,z_2)\right\|=|x_1|+|x_2|+2\left(\sqrt{y_1^2+z_1^2}+\sqrt{y_2^2+z_2^2}\right)\geq$$ $$\geq|x_1+x_2|+2\sqrt{(y_1+y_2)^2+(z_1+z_2)^2}=\left\|(x_1+x_2,y_1+y_2,z_1+z_2)\right\|.$$ Can you end it now?

Answer 2

Triangle in equality is obvious from the fact that $\|(x,y,z)\|=\|(x,0,0)\|'+2\|(0,y,z)\|'$ where $\|.\|'$ is the Euclidean norm.