How to prove a norm on three dimensional space Show that $||(x,y,z)||=|x| + 2 \sqrt{(y^2+z^2)}$ is a norm on $\mathbb R^3 $. Sketch the unit ball. I have to show the positiveness, homogeneity and triangle inequality of the equation. So notice that $|x| + x \sqrt{y^2+z^2)} = 0$ iff $ ||x || = 0$, $ ||y|| = 0$ and $ ||z|| = 0$.Hence it satifies the postive definiteness of the norm. Suppose $ \alpha \in \mathbb R$, then $$ ||\alpha (x,y,z)|| = \alpha [|x| + 2 \sqrt{(y^2+z^2)}] = |\alpha| ||(x,y,z)||$$ Hence it satisfies the nomogeneity. But I got stuck how to show triangle inequality for 3 variables. Also, I need help for sketching the unit ball. Please guide me to finish the problem. Thanks in advance!

For the triangle inequality, you have to prove: $$\lvert x+x'\rvert+2\sqrt{(y+y')^2+(z+z')^2}\leq\lvert x\rvert+2\sqrt{y^2+z^2}+\lvert x'\rvert+2\sqrt{y'^2+z'^2}$$ This amounts to proving: \begin{align*} \sqrt{(y+y')^2+(z+z')^2}&\leq \sqrt{y^2+z^2}+\sqrt{y'^2+z'^2}\\\ \iff(y+y')^2+(z+z')^2&\leq y^2+z^2+y'^2+z'^2+2 \sqrt{y^2+z^2}\sqrt{y'^2+z'^2}\\\ \iff\phantom{(y+y')^2+(z+z')^2}\llap{yy'+zz'}&\leq\sqrt{y^2+z^2}\sqrt{y'^2+z'^2} \end{align*} This is simply _Cauchy-Schwarz_ 's inequality.