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√(a²;p+b²;q)≥|ap+bq|證:由題q = 1-p(0≤p,q≤1)①|ap+bq|²;= a²;p²;+ b²;q²;+ 2abpq代入①= a²;p²;+ b²;(1-p)²;+2abp(1-p)=(a²;-2ab+b²;…
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