Question

已知a,b,c属于R+,求证：3[(a+b+c)/3-三次根号下（abc)]≥2[(a+b)/2-根号下(ab)]

Answer 1

原不等式整理后即证

c+2(ab)^(1/2)≥3(abc)^(1/3)

又由均值不等式知：

左边=c+(ab)^(1/2)+(ab)^(1/2)

≥3[c*(ab)^(1/2)*(ab)^(1/2)]

=3(abc)^(1/3)

=右边

得证