Proving triangle inequality of metric space Since we know that the hyperbolic plane $\mathbb{H^{2}}$ is metric space with metric $ d_{H}(z,w) = 2 tan h^{-1} \frac{ \mid z - w \mid}{ \mid z - \overline{w} \mid }$, where $ z, w \in \mathbb{H^{2}}$. Now we define the distance function on $\mathbb{H^{2}} \times \mathbb{H^{2}} $ given by $ \gamma$(z,w) = ( $ (d_{H}(z_{1},w_{1}))^{2} + (d_{H}(z_{2},w_{2}))^{2})^{\frac{1}{2}}$, where $ z = ( z_{1}, z_{2}) , w =  (w_{1}, w_{2})  \in \mathbb{H^{2}} \times \mathbb{H^{2}}$. I want to prove the triangle inequality of this $\gamma$. I tried to prove it by using Minkowski's inequality as it looks like $L_{2}$ metric,  but couldn't prove it. So can some one help me in carrying out the proof.
 A: Here are some more details to help you out.  Let's consider the function $f: [0,\infty)^2 \to [0,\infty)$ given by $f(a,b) = \sqrt{a^2 + b^2}$.  The usual triangle inequality in $2D$ Euclidean space implies that
$$
f(a_1 + a_2 , b_1 + b_2) \le f(a_1,b_1) + f(a_2,b_2).
$$
Indeed, the above is equivalent to 
$$
\Vert (a_1+a_2,b_1+b_2) \Vert \le \Vert (a_1,b_1) \Vert + \Vert (a_2,b_2) \Vert.
$$
Now we note another key property of $f$.  If $0 \le a_1 \le a_2$ and $0 \le b_2 \le b_2$, then 
$$
a_1^2 \le a_2^2 \text{ and } b_1^2 \le b_2^2 \Rightarrow a_1^2 + b_1^2 \le a_2^2 + b_2^2
$$
and so
$$
f(a_1,b_1) = \sqrt{a_1^2 + b_1^2} \le \sqrt{a_2^2 + b_2^2} = f(a_2,b_2).
$$
Now let $(X,d)$ and $(Y,\rho)$ be two metric spaces.  On $Z = X \times Y$ we define 
$$
\sigma((x_1,y_1),(x_2,y_2)) = f(d(x_1,x_2) , \rho(y_1,y_2) ).
$$
Let's prove the triangle inequality for $\sigma$.  Let $x_1,x_2,x_3 \in X$ and $y_1,y_2,y_3 \in Y$.  Then 
$$
d(x_1,x_2) \le d(x_1,x_3) + d(x_3,x_2) \text{ and } \rho(y_1,y_2) \le \rho(y_1,y_3) + \rho(y_3,y_2),
$$
and so the properties of $f$ above allows us to estimate
$$
\sigma((x_1,y_1),(x_2,y_2)) = f(d(x_1,x_2) , \rho(y_1,y_2) )  \\
\le f(d(x_1,x_3) + d(x_3,x_2) , \rho(y_1,y_3) + \rho(y_3,y_2) ) \\
\le f(d(x_1,x_3), \rho(y_1,y_3) ) + f(d(x_3,x_2), \rho(y_3,y_2)) \\
= \sigma((x_1,y_1), (x_3,y_3)) + \sigma((x_3,y_3),(x_2,y_2)).
$$
Thus the triangle inequality holds for $\sigma$.  Positivity and symmetry can be proved easily, so $\sigma$ is a metric on $X \times Y$.
