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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.

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    Perhaps the concreteness of the hyperbolic metric gets in the way? If you try to prove that for all metric spaces $(X,d_X)$ and $(Y,d_Y)$ the function $$d_p\bigl((x_1,y_1),(x_2,y_2)\bigr) = \sqrt{d_X(x_1,x_2)^2 + d_Y(y_1,y_2)^2}$$ defines a metric on $X\times Y$, you have only the general properties to work with, maybe that helps. – 2017-01-11
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    Actually, I also thought in that way, but I couldn't prove it. – 2017-01-11

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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 $X \times Y$.