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Define $$\|x\|=\sqrt[3]{(|x_1|^2+|x_2|^2)^{3/2}+|x_3|^3}$$ for any $x=(x_1,x_2,x_3)\in \Bbb R^3$. Show that $\|\cdot\|$ is a norm on $\Bbb R^3$.

I stuck on showing triangle inequality. I don't know how to show $$\|x+y\|\le \|x\|+\|y\|$$ for $x,y\in \Bbb R^3$. That is $$\sqrt[3]{(|x_1+y_1|^2+|x_2+y_2|^2)^{3/2}+|x_3+y_3|^3}\le \sqrt[3]{(|x_1|^2+|x_2|^2)^{3/2}+|x_3|^3}+\sqrt[3]{(|y_1|^2+|y_2|^2)^{3/2}+|y_3|^3}$$ How do I show that?

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Note, that $\|\cdot\|$ is somehow the composition of the $2$-norm and the $3$-norm on $\mathbb R^2$. We have $$\|x\| = \|(\|(x_1,x_2)\|_2,x_3)\|_3$$ and with this at hand the proof is easy (but a bit ugly).

We have by the triangle inequality for the $2$-norm

$$a:= \|(x_1+y_1,x_2+y_2)\|_2 \leq \|(x_1,x_2)\|_2 + \|(y_1,y_2)\|_2 =:b$$

and since for all $a,b\in\mathbb R$ the inequality $|a| \leq |b|$ induces $\|(a,x_3+y_3)\|_3 \leq \|(b,x_3+y_3)\|_3$, we obtain (by using also the triangle inequality for the $3$-norm)

$$\|x+y\| = \|(\|(x_1+y_1,x_2+y_2)\|_2,x_3+y_3)\|_3 \\ \leq \|(\|(x_1,x_2)\|_2 + \|(y_1,y_2)\|_2,x_3+y_3)\|_3\\ \leq \|(\|(x_1,x_2)\|_2,x_3)\|_3 + \|(\|(y_1,y_2)\|_2,y_3)\|_3\\ = \|x\| + \|y\|.$$

This procedure could of course be generalised.

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