已知a,b,c∈(0,+∞),且a+b+c=1,求证:(1/a)+(1/b)+(1/c)≥9.
热心网友
(1/a)+(1/b)+(1/c) = (a+b+c)/a +(a+b+c)/b +(a+b+c)/c= 3 +(b/a +c/a +c/b +a/b +a/c +/b/c)≥ 3 + 6*[(b/a)(c/a)(c/b)(a/b)(a/c)(/b/c)]^(1/6)= 3 + 6 = 9
已知a,b,c∈(0,+∞),且a+b+c=1,求证:(1/a)+(1/b)+(1/c)≥9.
(1/a)+(1/b)+(1/c) = (a+b+c)/a +(a+b+c)/b +(a+b+c)/c= 3 +(b/a +c/a +c/b +a/b +a/c +/b/c)≥ 3 + 6*[(b/a)(c/a)(c/b)(a/b)(a/c)(/b/c)]^(1/6)= 3 + 6 = 9