Question

实数x,y,z 满足x²+y²+z²=1,则根号2xy+yz的最大值是为？？？

Answer 1

x²+4y²/5≥2xy*2/根号5,z²+y²/5≥yz*2/根号5
(2xy+yz)*2/根号5≤1，2xy+yz≤根号5/2。当且仅当x²=4y²/5且z²=y²/5时等式成立，此时x²=2/5,y²=1/2,z²=1/10

Answer 2

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
7月V8

追问

这什么呀，a哪来的啊