Show that $d(x,y)={2|x-y|\over \sqrt{1+|x^2|}\sqrt{1+|y^2|}}$ is a metric on $\mathbb C$

Question

Prove that the function $d:\mathbb{C×C}\rightarrow \mathbb R$ defined by $$d(x,y)={2|x-y|\over \sqrt{1+|x^2|}\sqrt{1+|y^2|}}$$ is a metric on $\mathbb C$.

My professor gave me this problem. The first three properties were easy to prove. But I couldn't prove the Triangle inequality. Just prove that part.

Answer 1

$\mathbf{Hint:}$ $1.\,\,|1+\bar xy|\le \sqrt{1+|x^2|}\sqrt{1+|y^2|} \,\,\forall x,y\in \mathbb C. \text{Start from} (|x|-|y|)^2\ge 0$ $2.$ $(x-z)(1+y\bar y)=(x-y)(1+\bar yz)+(y-z)(1+\bar yx) \,\,\forall x,y,z \in \mathbb C$. Just simplify the RHS & LHS.

Then take modulus on both sides of (2).