The title says it all ... more formally : let $n \geq 1$, and let $a_1, a_2 , \ldots ,a_n$ be positive numbers, let $b_1, b_2 , \ldots ,b_n$ be real numbers. Consider for $x\in {\mathbb R}$,
$ \phi(x)=\sum_{k=1}^n a_k |x-b_k| $
It is easy to see that $\phi$ has a global minimum $m$ on $\mathbb R$, and that $m$ is attained of one the values $b_1, \ldots ,b_n$. In fact, $\phi$ is affine on each interval $[b_k,b_{k+1}]$ (and on the outer intervals $]-\infty,b_1]$ and $[b_n,+\infty[$ ).
I say that "m is the global minimum for $\phi$" can be proved "using only the triangular inequality" if there are numbers a'_1,a'_2, \ldots ,a'_n with 0 \leq a'_k \leq a_k (for $1 \leq k \leq n$) and constants $\varepsilon_1,\varepsilon_2, \ldots, \varepsilon_n$ all equal to $-1$ or $+1$, such that
\sum_{k=1}^n\varepsilon_ka'_k=0, \ {\rm and} \ \sum_{k=1}^n\varepsilon_ka'_kb_k=m
because then we have
\phi(x)=\sum_{k=1}^n a_k |x-b_k| \geq \sum_{k=1}^n a'_k |x-b_k| =\sum_{k=1}^n |\varepsilon_k a'_k(x-b_k)| \geq \Bigg| \sum_{k=1}^n \varepsilon_k a'_k(x-b_k) \Bigg| =m