实数x,y,z满足x^2+y^2+z^2=1,则sqr(2)xy+yz的最大值为
发布网友
发布时间：2024-10-05 09:21
共2个回答
热心网友
时间：2024-10-05 10:00
>=2*sqrt(sqrt(3)/2*x^2*sqrt(3)/3*y^2)
=2sqrt(1/2*x^2y^2)=sqrt(2)xy -----(1)
sqrt(3)/6*y^2+sqrt(3)/2*z^2
>=2*sqrt(sqrt(3)/6*y^2*sqrt(3)/2*z^2)
=2sqrt(1/4y^2*z^2)=yz ------(2)
=>(1)+(2)
=>sqrt(2)xy+yz<=sqrt(3)/2*x^2+sqrt(3)/3*y^2+sqrt(3)/6*y^2+sqrt(3)/2*z^2=sqrt(3)/2(x^2+y^2+z^2)=sqrt(3)/2
故sqr(2)xy+yz的最大值为为sqrt(3)/2
当且仅当sqrt(3)/2*x^2=sqrt(3)/3*y^2，sqrt(3)/6*y^2=sqrt(3)/2*z^2
x^2+y^2+z^2=1=>x=sqrt(3)/3 y=sqrt(2)/2 z=sqrt(6)/6
热心网友
时间：2024-10-05 10:03
>=2*sqrt(sqrt(3)/2*x^2*sqrt(3)/3*y^2)
=2sqrt(1/2*x^2y^2)=sqrt(2)xy
-----(1)
sqrt(3)/6*y^2+sqrt(3)/2*z^2
>=2*sqrt(sqrt(3)/6*y^2*sqrt(3)/2*z^2)
=2sqrt(1/4y^2*z^2)=yz
------(2)
=>(1)+(2)
=>sqrt(2)xy+yz<=sqrt(3)/2*x^2+sqrt(3)/3*y^2+sqrt(3)/6*y^2+sqrt(3)/2*z^2=sqrt(3)/2(x^2+y^2+z^2)=sqrt(3)/2
故sqr(2)xy+yz的最大值为为sqrt(3)/2
当且仅当sqrt(3)/2*x^2=sqrt(3)/3*y^2，sqrt(3)/6*y^2=sqrt(3)/2*z^2
x^2+y^2+z^2=1=>x=sqrt(3)/3
y=sqrt(2)/2
z=sqrt(6)/6
