Question

实数x,y,z 满足x²+y²+z²=1,则根号2xy+yz的最大值是多少

过程

Answer 1

yz<=1/(4a)*y^2+a*z^2
(根号2)xy<=a*x^2+1/(2a)*y^2

so
(根号2)xy+yz<=a*x^2+1/(2a)*y^2+1/(4a)*y^2+a*z^2
=a*x^2+3/(4a)*y^2+a*z^2

令a=3/(4a)
即a=根号3/2
那么(根号2)xy+yz<=a*x^2+1/(2a)*y^2+1/(4a)*y^2+a*z^2
=a*x^2+3/(4a)*y^2+a*z^2=a=根号3/2