已知a.b,c是正整数，且互不相等，且abc=1求证：√a+√b+√c＜１／a+1／b+1／c
热心网友
因为abc=1,所以√a+√b+√c=√(1/bc)+√(1/ac)+√(1/ab)........而1/a+1/b+1/c=(1/2a+1/2b)+(1/2b+1/2c)+(1/2a+1/2c)(1/2a+1/2b)＞2√(1/2a)(1/2b)=√(1/ab)..........(1/2b+1/2c)＞2√(1/2b)(1/2c)=√(1/bc)..........(1/2a+1/2c)＞2√(1/2a)(1/2c)=√(1/ac)..........++得:1/a+1/b+1/c＞√(1/ab)+√(1/bc)+√(1/ac)=即：√a+√b+√c＜１／a+1／b+1／c
热心网友
题目错误，a.b,c是正整数，且互不相等，且abc=1明显自相矛盾！