- 设x、y、z∈R+,√(x^2+y^2)+z=1,求xy+2xz?
- ∵√(x^2+y^2)+z=1且x、y、z∈R+,
故可令z=(sinα)^2,√(x^2+y^2)=(cosα)^2,
而x=(cosα)^2sinβ,y=(cosα)^2cosβ,
其中,α、β∈(0,π/2].
于是,代入所求式整理,得
xy+2xz
=[sinβ/(2-cosβ),2(cosα)^2-cosβ(cosα)^2,cosβ(cosα)^2+2(sinα)^2]
≤[sinβ/(2-cosβ),(2(cosα)^2-cosβ(cosα)^2+cosβ(cosα)^2+2(sinα)^2)/2]^2
=sinβ/(2-cosβ).
令tan(β/2)=t,则依万能代换公式有,
xy+2xz
≤[2t/(1+t^2)]/[2-((1-t^2)/(1+t^2))]
=2t/(1+3t^2)
≤2t/(2√3t)
=√3/3.
当x=√3/3,y=z=1/3时,上式取等号.
即xy+2xz的最大值为√3/3。