# Proving metric( the triangle inequality)

In complex plane $\mathbb{C}$, how to prove that $$d(z_1,z_2)=\frac{|z_2-z_1|}{\sqrt{1+|z_1|^2}\sqrt{1+|z_2|^2}}$$ is a metric?

I got stuck in the triangle inequality, and have no idea of proving it, any hint or solutions?

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You decribe the def. of $d$ correctly? I think this $d$ is not a metric, because $d(z_1,z_2)$ can be nonpositive real number (or even nonreal complex number.) –  tetori Jan 23 '13 at 10:27
Do you mean $\lvert z_1 \rvert^2$? otherwise, e.g. for $z_1=i$, this isn't well-defined. –  k.stm Jan 23 '13 at 10:27
Oh yes， I mean $|z|^2$,sorry for my mistake. –  Golbez Jan 23 '13 at 11:06