$(X, d_X)$ and $(Y, d_Y)$ are metric spaces and $Z = X ×Y$ the cartesian product of $X$ and $Y$.
$d_Z: Z × Z \to ℝ_\geq0$ and $d_Z:=\sqrt{d_X^2+d_Y^2} $
So this is:
$d_Z((x,y),(x',y')) = \sqrt{d_X(x,x')^2+d_Y(y,y')^2}$
How to prove the triangle inequality for showing $(Z,d_Z)$ is a metric space?
I tried with
$d_Z((x,x'),(y,y'))+d_Z((y,y'),(z,z')) \geq d_Z((x,x'),(z,z'))$
So i put the definition in:
$\sqrt{d_X(x,y)^2+d_Y(x',y')^2}+\sqrt{d_X(y,z)^2+d_Y(y',z')^2}$
But i don't know what to do next to prove that this is $\geq$ than
$\sqrt{d_X(x,z)^2+d_Y(x',z')^2}=d_Z((x,x'),(z,z'))$
Note that
$$d_Z((x,y),(x',y'))=\lVert\big(d_X(x,x'), d_Y(y,y')\big)\rVert$$
where $\lVert\cdot\rVert$ is the Euclidean norm in $\mathbb{R}^2$. Then we have:
$$d_Z\big((x,y),(x'', y'')\big)+d_Z\big((x'', y''),(x',y')\big)=$$ $$=\lVert\big(d_X(x,x''), d_Y(y,y'')\big)\rVert+\lVert\big(d_X(x'',x'), d_Y(y'',y')\big)\rVert\geq$$ $$\geq\lVert\big(d_X(x,x''), d_Y(y,y'')\big) + \big(d_X(x'',x'), d_Y(y'',y')\big)\rVert=$$ $$=\lVert\big(d_X(x,x'')+d_X(x'', x'), d_Y(y,y'')+d_Y(y'',y')\big)\rVert\geq$$ $$\geq \lVert\big(d_X(x,x'), d_Y(y,y')\big)\rVert=$$ $$d_Z\big((x,y), (x',y') \big)$$
The first inequality is because every norm is subadditive. The second one because $d_X$ and $d_Y$ satisfy the triangle inequality.