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1/(ab)+1/a(a-b)=(1/a)[1/b+1/(a-b)]=(1/a)[(a-b+b)/b(a-b)]=1/b(a-b) 因為b(a-b)≤[(b+a-b)/2]²;=a²;/4 所以1/b(a-b)≥4/a²; 即1/(ab)+1/a(a-b)≥4/a²; 注:考慮一下,條件應為a>b>0
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