Aratati ca oricare ar fi numerele complexe a, b, c are loc inegalitatea
\( |a|+|b|+|c|\leq|a+b+c|+|a-b|+|b-c|+|c-a| \)
Mihai Opincariu G.M.B. 2000
Inegalitate cu numere complexe
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Folosind inegalitatile cunoscute cu modul avem:
\( |a+b|+|a-b|\ge |a+b+a-b|=2|a| \)
\( |a+b|+|a-b|\ge |a+b-(a-b)|=2|b| \)
\( \Longrightarrow |a+b|+|a-b|\ge |a|+|b| \)
Sumand si inegalitatile analoage obtinem:
\( |a+b|+|b+c|+|c+a|+|a-b|+|b-c|+|c-a|\ge 2(|a|+|b|+|c|) \ (1) \)
Din inegalitatea lui Hlawka avem:
\( |a+b+c|+|a|+|b|+|c|\ge |a+b|+|b+c|+|c+a|\ (2) \)
Adunand relatiile \( (1) \) si \( (2) \) obtinem concluzia.
\( |a+b|+|a-b|\ge |a+b+a-b|=2|a| \)
\( |a+b|+|a-b|\ge |a+b-(a-b)|=2|b| \)
\( \Longrightarrow |a+b|+|a-b|\ge |a|+|b| \)
Sumand si inegalitatile analoage obtinem:
\( |a+b|+|b+c|+|c+a|+|a-b|+|b-c|+|c-a|\ge 2(|a|+|b|+|c|) \ (1) \)
Din inegalitatea lui Hlawka avem:
\( |a+b+c|+|a|+|b|+|c|\ge |a+b|+|b+c|+|c+a|\ (2) \)
Adunand relatiile \( (1) \) si \( (2) \) obtinem concluzia.