已知x，y，z∈R，且x+y+z=1，求证：1x+4y+9z≥36．

Question

已知x，y，z∈R，且x+y+z=1，求证：(1)/(x)+(4)/(y)+(9)/(z)≥36．

Answer 1

证明：(1)/(x)+(4)/(y)+(9)/(z)=（(1)/(x)+(4)/(y)+(9)/(z)）（x+y+z）=14+(y)/(x)+(4x)/(y)+(z)/(x)+(9x)/(z)+(4z)/(y)+(9y)/(z)≥14+4+6+12=36，
当且仅当(y)/(x)=(4x)/(y)，(z)/(x)=(9x)/(z)，(4z)/(y)=(9y)/(z)时，等号成立．