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This question already has an answer here:

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?

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Hint:

$$(1-a^2)(1-b^2)>0$$

if $1-a^2,1-b^2>0\iff a^2,b^2<1\iff-1<a,b<1$

or if both $<0$

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