设x，y，z是正实数，且xyz=1．证明：x3(1+y)(1+z)+y3(1+z)(1+x)+z3(1+x)(1+y)≥34．

Question

设x，y，z是正实数，且xyz=1．
证明：(x3)/((1+y)(1+z))+(y3)/((1+z)(1+x))+(z3)/((1+x)(1+y))≥(3)/(4)．

Answer 1

证明：根据均值不等式得：
(x3)/((1+y)(1+z))+(1+y)/(8)+(1+z)/(8)≥(3x)/(4)①
(y3)/((1+z)(1+x))+(1+x)/(8)+(1+z)/(8)≥(3y)/(4)②
(z3)/((1+x)(1+y))+(1+x)/(8)+(1+y)/(8)≥(3z)/(4)③
①+②+③得
(x3)/((1+y)(1+z))+(y3)/((1+z)(1+x))+(z3)/((1+x)(1+y))≥(x+y+z)/(2)−(3)/(4)
∵x+y+z≥3

3

xyz

=3
∴(x3)/((1+y)(1+z))+(y3)/((1+z)(1+x))+(z3)/((1+x)(1+y))≥(3)/(4)