If $\mathcal{M}=\{M_i : i\in I_n\}$ is a collection of metric spaces, each with metric $d_i$, we can make $M=\prod_{i\in I_n}M_i$ a metric space using the $p$-norm, we simply set $d : M\times M\to \mathbb{R}$ as:
$$d((p_1,\dots,p_n),(q_1,\dots,q_n))=\left\|(d(p_1,q_1),\dots,d(p_n,q_n))\right\|_p$$
What I want to prove is that the $p$-norm
$$\left\|x\right\|_p=\left(\sum_{i=1}^{n}\left|x_i\right|^p\right)^{1/p}$$
is really a norm. Showing that $\left\|x\right\|_p \geq 0$ being zero if and only if $x = 0$ was easy. Showing that $\left\|kx\right\|_p = \left|k\right|\left\|x\right\|_p$ was also easy. The triangle inequality is the thing that is not being easy to show. Indeed, I want to show that: for every $x,y \in \mathbb{R}^n$ we have:
$$\left(\sum_{i=1}^{n}\left|x_i+y_i\right|^p\right)^{1/p}\leq \left(\sum_{i=1}^{n}\left|x_i\right|^p\right)^{1/p}+\left(\sum_{i=1}^{n}\left|y_i\right|^p\right)^{1/p}.$$
I thought that it might not be as difficult as it seems, but after trying a little without sucess I've searched on the internet and the I found that we need measure theory to prove that. Is there any more elementary proof of this inequality?