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问题: 急

求证;1/a+1/b+1/c大于根号下1/ab+1/bc+1/ac a,b,c为正数,且互不相等。求a b c?

解答:

(1/a+1/b+1/c)^2≥0
(1/a)^2+(1/b)^2+(1/c)^2+2/ab+2/bc+2/ac≥0
(1/a)^2+(1/b)^2+(1/c)^2+2/ab+2/bc+2/ac≥1/ab+1/bc+1/ac(a,b,c为正数)
所以(1/a+1/b+1/c)^2>1/ab+1/bc+1/ac
所以1/a+1/b+1/c>√1/ab+1/bc+1/ac
不知道是否恰当,共参考!