How to prove triangle inequality for $p$-norm?
If you mean $$ \left\|x\right\|_p=\left(\sum_{i=1}^n\left|x_i\right|^{\color{#C00000}{p}}\right)^{1/p}\tag{1} $$ then Minkowski's Inequality is the triangle inequality for the $p$-norm.
Duality
Note that by Hölder's Inequality, if $\|y\|_q=1$, where $\frac1p+\frac1q=1$, we have $$ \left|\sum_{i=1}^nx_iy_i\right|\le\|x\|_p\tag{2} $$ Furthermore, if $y_i=\frac{\bar{x}_i|x_i|^{p/q-1}}{\|x\|_p^{p/q}}$, then $\|y\|_q=1$ and $$ \sum_{i=1}^nx_iy_i=\|x\|_p\tag{3} $$ $(2)$ and $(3)$ show that $$ \|x\|_p=\sup_{\|y\|_q=1}\left|\sum_{i=1}^nx_iy_i\right|\tag{4} $$ $(4)$ says that $\ell^p$ is the dual of $\ell^q$.
Duality Proof of Minkowski's Inequality $$ \|x+y\|_p =\sup_{\substack{\|u\|_q&=1\\\|v\|_q&=1\\u&=v}}\sum_{i=1}^nx_iu_i+y_iv_i \le\sup_{\substack{\|u\|_q&=1\\\|v\|_q&=1}}\sum_{i=1}^nx_iu_i+y_iv_i =\|x\|_p+\|y\|_p\tag{5} $$ The inequality is because the $\sup$ on the left is being taken over a subset of the pairs $(u,v)$ over which the $\sup$ on the right is being taken.
(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.