已知:abc=1,a>0,b>0,c>0,求证:(b+c)/a+(a+c)/b+(a+b)/c>=2(a+b+c)

题目

已知:abc=1,a>0,b>0,c>0,求证:(b+c)/a+(a+c)/b+(a+b)/c>=2(a+b+c)

答案

2a/b+2a/c+b/c+c/b
= a/b+a/b+a/c+a/c+b/c+c/b
≥ 6(a/b·a/b·a/c·a/c·b/c·c/b)^(1/6) (均值不等式)
= 6(a^4/(bc)^2)^(1/6)
= 6(a^6/(abc)^2)^(1/6)
= 6a (abc = 1).
同理, 2b/c+2b/a+c/a+a/c ≥ 6b, 2c/a+2c/b+a/b+b/a ≥ 6c.
相加得3(a/b+a/c+b/c+b/a+c/a+c/b) ≥ 6(a+b+c).
即(a+b)/c+(b+c)/a+(c+a)/b ≥ 2(a+b+c).