Fie \( a, b \) doua numere complexe. Sa se demonstreze ca \( |1+ab|+|a+b|\geq \sqrt{|a^2-1|\cdot|b^2-1|} \).
***, Olimpiada judeteana 2008
Inegalitate cu module de numere complexe
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Problema e chiar simpla, desi nu m-am prins din prima:
\( |1+ab|+|a+b|=\frac{|1+ab|+|a+b|+|1+ab|+|-a-b|}{2}\geq \frac{|(a+1)(b+1)|+|(a-1)(b-1)|}{2}\geq \sqrt{|a^2-1||b^2-1|} \).
Am aplicat medii o data si inegalitatea modulului de 2 ori
\( |1+ab|+|a+b|=\frac{|1+ab|+|a+b|+|1+ab|+|-a-b|}{2}\geq \frac{|(a+1)(b+1)|+|(a-1)(b-1)|}{2}\geq \sqrt{|a^2-1||b^2-1|} \).
Am aplicat medii o data si inegalitatea modulului de 2 ori
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