Question

已知x²+y²+z²=1,则x+2y+3z的最大值是…(　　)

A.5                 B.50                      C.                    D.

Answer 1

解析:根据柯西不等式,有?

(x+2y+3z)²≤(x²+y²+z²)(1²+2²+3²)=14,?

∴x+2y+3z≤.∴选D.?

答案：D