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首先,補充一下,這個不等式一定要是a>1,b>1,c>1才可以成立的. 將原方程化為: a²;+b²;+c²;-ab-bc-ac≥0 2a²;+2b²;+2c²;-2ab-2bc-2ac≥0 (a-b)²;+(b-c)²;+(a-c)²;≥0 又∵(a-b)²;≥0,(b-c)²;≥0,(a-c)²;≥0, ∴原不等式成立. C式工作室為您解答,
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