If $|a|,|b|<1$, prove that $\frac{|a+b|}{|1+ab|}<1$.

Question

So I've gotten as far as $\displaystyle\frac{|a+b|}{|1+ab|}<\frac{2}{|1+ab|}$ which is clearly wrong because it is greater than 1. What am I doing wrong? Is this question even true?

Note: computing a weaker bound, such as you did, is not wrong -- merely insufficient. — vadim123, Mar 24 '14 at 18:40

score 1 · Answer 1 · answered Mar 24 '14 at 18:43

1

We need to prove: /a + b/ < /1 + ab/ <==> (a + b)^2 < (1 + ab)^2 <==> a^2 + 2ab + b^2 < 1 + 2ab + a^2b^2 <==>

(1 - a^2)(1 - b^2) > 0 which is true since /a/ < 1 and /b/ < 1.

answered Mar 24 '14 at 18:43

DeepSea

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If $|a|,|b|<1$, prove that $\frac{|a+b|}{|1+ab|}<1$.

1 Answers1