# $a^2+b^2<1+a^2b^2$ [duplicate]

If $$a<1$$ and $$b<1$$ then how do I prove that $$a^2+b^2<1+a^2b^2$$

I stumble across this equation while solving a problem from complex analysis i.e. $$|a-b|/|1-(\bar{a})b|<1 \ \mbox{if } |a|<1 \mbox{ and } |b|<1$$ where $$\bar{a}$$ = conjugate of a. can we prove second without proving first?

## marked as duplicate by Martin R, Community♦Jan 11 at 12:54

$$(1-a^2)(1-b^2)>0$$
if $$1-a^2,1-b^2>0\iff a^2,b^2<1\iff-1
or if both $$<0$$