How to Prove a Complex Inequality with Complex Algebra

AI Thread Summary
The discussion revolves around proving the inequality |1+ab| + |a+b| ≥ √(|a²-1||b²-1|) for complex numbers a and b. Participants suggest factoring the left-hand side and exploring the expression |(a-1)(a+1)(b-1)(b+1)| to find a solution. The approach includes rearranging and recombining factors to simplify the proof. There is a focus on utilizing properties of complex algebra to tackle the problem. Overall, the thread emphasizes collaborative problem-solving in complex inequalities.
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Homework Statement


Let b and a be two complex numbers. Prove that
|1+ab| + |a + b| ≥ √(|a²-1||b²-1|).

Homework Equations


Complex algebra

The Attempt at a Solution


I don't know how to proceed. I posted it here to get some ideas :p
 
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Try factoring the left hand side:
$$|a^2 - 1||b^2 - 1| = |(a-1)(a+1)(b-1)(b+1)|$$
Now see what happens if you rearrange the factors and recombine them.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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