Prove that $||x||= [(|\alpha_{1}|^2+|\alpha_{2}|^3)^{3/2} + |\alpha_{3}|^3]^{1/3}$ defines a norm Let the vector space $X=K^3$. For $x=(\alpha_{1}, \alpha_{2}, \alpha_{3}) \in X$, we define
$||x||= [(|\alpha_{1}|^2+|\alpha_{2}|^3)^\frac{3}{2} + |\alpha_{3}|^3]^\frac{1}{3}$ 
Proof that $||·||$ is a norm.
To prove that it is a norm,  I have to proof 4 properties. I know how to prove 3 of them but I don´t know how to prove the triangle inequality, i.e:
$||x+y||≤||x||+||y||$
If you can help me please. Thank you very much.
 A: The exercise probably asks to consider, for $x=(u,v,w)$,
$$\|x\|= [(|u|^2+|v|^{\color{red}{\bf 2}})^{3/2} + |w|^3]^{1/3}.
$$ 
(Note the wrong exponent in the question.) To show that this identity indeed defines a norm, let us enlarge somewhat the setting and consider, for every $p\geqslant1$, the norm $N_p$ defined on $\mathbb R^2$ by
$$
N_p(u,v)=(|u|^p+|v|^p)^{1/p}.
$$
Then,
$$
\|x\|= N_3(N_2(u,v),w),
$$
thus, it suffices to show the following:

If $N$ and $M$ are norms on $\mathbb R^2$, then $Q$ defined by
  $
Q(u,v,w)=M(N(u,v),w),
$
  is a norm on $\mathbb R^3$. 

Let us concentrate on the triangular inequality. One has
$$
Q(u+u',v+v',w+w')=M(N(u+u',v+v'),w+w'),
$$
and the triangular inequality for $N$ yields
$$
N(u+u',v+v')\leqslant N(u,v)+N(u',v').
$$
Assume that $M$ satisfies the following property:

$(\ast)$ For every $t$, the function $x\mapsto M(x,t)$ is nondecreasing on $x\geqslant0$.

Then
$$
M(N(u+u',v+v'),w+w')\leqslant M(N(u,v)+N(u',v'),w+w').
$$ 
The triangular inequality for $M$ yields that the RHS is at most
$$
M(N(u,v),w)+M(N(u',v'),w')=Q(u,v,w)+Q(u',v',w'),
$$
which proves that $Q$ satisfies the triangular inequality. Every norm $N_p$ satisfies $(\ast)$ hence the question asked is solved and the method actually shows that, for every $n\geqslant1$ and $m\geqslant1$,
$$\|x\|= [(|u|^n+|v|^n)^{m/n} + |w|^m]^{1/m},
$$
defines a norm.
Caveat: We used that $N$ and $M$ are norms, and the additional property that, for every $t$, $x\mapsto M(x,t)$ is nondecreasing on $x\geqslant0$. Unless I am missing something, to determine whether one can get rid of this hypothesis or not, is an interesting question.
