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(a+b+c)²;=a²;+b²;+c²;+2ac+2ab+2bc=a²;+b²;+c²;+2(ab+ac+bc) 又因為a>0 b>0 c>0 所以得到a²;+b²;+c²;>0 由(a+b+c)²;-2(ab+ac+bc)=a²;+b²;+c²;>0 所以得到(a+b+c)²;>2(ab+ac+bc)
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