Q:

Metric space $M = M_1\times \cdots M_n$ with metric $d''(z,z') = \max \{d_i(x_i,x_i'), i\in \{1,\cdots,n\}\}$. Showing triangular inequality.

I have the following metric space:
$M = M_1\times \cdots \times M_n$ with metric $d''(z,z') = \max \{d_i(x_i,x_i'), i\in \{1,\cdots,n\}\}$
where $d_i$ is the metric for each $M_i$, and $z=(x_1,\cdots,x_n), z'=(x_1',\cdots,x_n')$
I already showed the other properties but I need to show the property of triangular inequality for this one.
But since each metric is unique for each $M_i$, I get confused in how to prove this inequality, since I don't even know the metrics, how can I suppose that some metric is less than another?

A:

Hint: use $max(a_1,...,a_n)+max(b_1,...,b_n)\geq max(a_1+b_1,...,a_n+b_n)$.

A:

Take $z,z',z''\in M$. You have to prove that
$$d''(z,z')\leq d''(z,z'')+d''(z'',z')$$
Notice that there exists $i_0\in\{1,...,n\}$ such that $d''(z,z')=d_{i_0}(x_{i_0},x_{i_0}')$. Thus, since $d_{i_0}$ is a metric,
$$\begin{align}
d''(z,z')&=d_{i_0}(x_{i_0},x_{i_0}')\\\\
&\leq d_{i_0}(x_{i_0}',x_{i_0}'')+d_{i_0}(x_{i_0}'',x_{i_0}')\\\\
&\leq \max \{d_i(x_i,x_i''), i\in \{1,\cdots,n\}\}+\max \{d_i(x_i'',x_i'), i\in \{1,\cdots,n\}\}\\\\
&=d''(z,z'')+d''(z'',z')
\end{align}$$

