How to show $d(x,y)= \sqrt{d_1(x_1,y_1)^2+d_2(x_2,y_2)^2}$ is a metric? $d_1(x_1,y_1)$ and  $d_2(x_2,y_2)$ are metric on $X$ and $d(x,y)$ is defined as:
$$d(x,y)= \sqrt{d_1(x_1,y_1)^2+d_2(x_2,y_2)^2}.$$
I am trying to show this is a metric. Can you give me some clue about proving the triangle inequality for $d$ ? Thank you for your help.
 A: Consider the function $f(a,b) = \sqrt{a^2 + b^2}$ on $[0,\infty)\times [0,\infty)$. Then 


*

*$f(a,b) \le f(c,d)$ whenever $a \le c$ and $b\le d$, and

*$f(a + c,b + d) \le f(a,b) + f(c,d)$,
where Property $2$ follows from the triangle inequality for vectors on $\Bbb R^2$. Given $x = (x_1,x_2)$, $y = (y_1,y_2)$, and $z = (z_1,z_2)$ in $X^2$,
\begin{align}
d(x,z) &= f(d_1(x_1,z_1), d_2(x_2,z_2))\\
&\le f(d_1(x_1,y_1) + d_1(y_1,z_1),d_2(x_2,y_2) + d_2(y_2,z_2)) \tag{i}\\
&\le f(d_1(x_1,y_1),d_2(x_2,y_2)) + f(d_1(y_1,z_1),d_2(y_2,z_2) \tag{ii}\\
&= d(x,y) + d(y,z)
\end{align}
Here, $(i)$ follows from Property 1. and the triangle inequalities for $d_1$ and $d_2$. Using the Property 2., we obtain (ii). Therefore, $d$ satisfies the triangle inequality.
A: \begin{align*}
d(x,z)&=\left(\sum d_i(x_i,z_i)^2\right)^{1/2}\leq\left(\sum(d_i(x_i,y_i)+d_i(y_i,z_i))^2\right)^{1/2}\\
&=\left(\sum d_i(x_i,y_i)^2+2\sum d_i(x_i,y_i)d_i(y_i,z_i)+\sum d_i(y_i,z_i)^2\right)^{1/2}\\
&\leq\left[\sum d_i(x_i,y_i)^2+2\left(\sum d_i(x_i,y_i)^2\right)^{1/2}\left(\sum d_i(y_i,z_i)^2\right)^{1/2}+\sum d_i(y_i,z_i)^2\right]^{1/2}\\
&=\left\{\left[\left(\sum d_i(x_i,y_i)^2\right)^{1/2}+\left(\sum d_i(y_i,z_i)^2\right)^{1/2}\right]^2\right\}^{1/2}=\left(\sum d_i(x_i,y_i)^2\right)^{1/2}+\left(\sum d_i(y_i,z_i)^2\right)^{1/2}\\
&\leq d(x,y)+d(y,z)
\end{align*}
