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|1 + ab| + |a + b| >= \sqrt{|a^2 - 1| \cdot |b^2 - 1|} , complex

Source: Indian Postal Coaching 2008 set 6 p3

May 25, 2020

complexinequalitiesalgebra

Problem Statement

Let

a

and

b

be two complex numbers. Prove the inequality

|1 + ab| + |a + b| \ge \sqrt{|a^2 - 1| \cdot |b^2 - 1|}