TITLE: How to prove triangle inequality for $p$-norm?
QUESTION [30 upvotes]: 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?

REPLY [21 votes]: (I learned from Terry Tao the following proof, which exploits a symmetry to simplify the task of proving an estimate by normalising one or more inconvenient factors to equal $1$.)
I assume here that $1\leq p<\infty$. We want to show that
$$
\|x+y\|_p\leq\|x\|_p+\|y\|_p\tag{*}
$$
When the RHS is $0$, the proof is trivial. Suppose it is positive. By homogeneity $\|cx\|_p=|c|\|x\|_p$ we may reduce to the case $\|x\|_p=1-\lambda$ and $\|y\|_p=\lambda$ for some $0\leq\lambda\leq 1$. The cases $\lambda=0,1$ are trivial, so suppose $0<\lambda<1$. Writing $X:=x/(1-\lambda)$ and $Y:=y/\lambda$ we reduce to the convexity estimate:
$$
\|(1-\lambda)X+\lambda Y\|_p\leq 1\quad\text{whenever } \|X\|_p=\|Y\|_p=1\
\text{and }0\leq\lambda\leq 1.
$$
But since $z\mapsto|z|^p$ is convex for $p\geq 1$, we have the coordinate-wise convexity bound
$$
|(1-\lambda)X_i+\lambda Y_i|^p\leq (1-\lambda)|X_i|^p+\lambda |Y_i|^p.
$$
Summing $i$ from $1$ to $n$, we obtain 
$$
\|(1-\lambda)X+\lambda Y\|_p^p\leq 1
$$
and thus the claim follows. 
Note that this proof works for the general abstract $L^p$ spaces as well.