已知2x+3y+4z=10，求x2+y2+z2的最小值．

Question

已知2x+3y+4z=10，求x²+y²+z²的最小值．

Answer 1

法1：∵2x+3y+4z=10，
∴x=5-(3)/(2)y-2x．
∴x²+y²+z²
=(5-(3)/(2)y-2z)2+y2+z2
=(13)/(4)y2+5z2+6zy-15y-20x+25
=(13)/(4)y2+(6z-15)y+5z2-20z+25
=(13)/(4)[y+(2(6z-15))/(13)]2+(29)/(13)z2-(80)/(13)z+(100)/(13)
=(13)/(4)(y+(12z-30)/(13))2+(29)/(13)(z-(40)/(29))2+(100)/(29)
≥(100)/(29)．
法2：由柯西不等式可得，（2x+3y+4z）²≤（x²+y²+z²）（2²+3²+4²），
由条件可得，x²+y²+z²≥(100)/(29)．
 故最小值为(100)/(29)．