Question

a>0,b>0,a3+b3(三次方）=2,证明：a+b

Answer 1

整理：sugar0763 - 魔法师 四级

2=a^3+b^3>=2(ab)^(3/2) 整理得 ab=-3

2=a^3+b^3=(a+b)(a^2-ab+b^2)=(a+b)[(a+b)^2-3ab]>=(a+b)[(a+b)^2-3]

==>(a+b)^3-3(a+b)-2[(a+b)^3+1]-[3(a+b)+1][(a+b)+1][(a+b)^2-(a+b)-2][(a+b)+1]{[(a+b)+1][(a+b)-2]}[(a+b)+1]^2[(a+b)-2]