Prove $|a+b-1| > |ab|$ where $|a|>1$, $a \in \mathbb{C}$, and $b \in (0,1)$.

Question

It may be stupid but I really do not know how to prove it.

Let $a \in \mathbb{C}$, and $|a| >1$. Let $b \in (0,1)$. Prove \begin{equation*} |a+b-1| > |ab|. \end{equation*}

Any help will be appreciated.

Answer 1

By AM-GM $$|a|^2+1> 2|a|\geq 2\Re(a).$$ Therefore $$|a|^2-1> 2\big(\Re(a)-1\big).$$ Since $|a|>1$, $|a|^2-1>0$. Thus $$(1+b)\big(|a|^2-1\big)>|a|^2-1>2\big(\Re(a)-1\big).$$ Thus $$(1-b^2)\big(|a|^2-1\big)=(1-b)\Big((1+b)\big(|a|^2-1\big)\Big)>2(1-b)\big(\Re(a)-1\big).$$ Hence $$|a|^2+b^2-|a|^2b^2-1>2\Re(a)+2b-2-2\Re(a)b,$$ or $$|a|^2+b^2+1-2b-2\Re(a)-2\Re(a)b>|a|^2b^2.$$ That is, $$|a+b-1|^2>|a|^2b^2=|ab|^2.$$

Answer 2

Let $a=x+iy$ with $x,y$ real, $x^2+y^2-1>0$ and $0<b<1$. Then $$|a+b-1|>|ab|$$ $$\Leftrightarrow (x+b-1)^2+y^2>b^2(x^2+y^2)~{\rm (by~raising~to~square)}$$ $$\Leftrightarrow (1-b)[(x-1)^2+b(x^2+y^2-1)+y^2]>0~({\rm by~rearranging}),$$ whence the result.