Question

已知x＞0，y＞0，x+y=1, 则x^2/(x+2)+y^2/(y+1)的最小值为

Answer 1

这是一个用判别式法，求分式函数值域的类型。

详情如图所示:

最小值为:1/4.

供参考，请笑纳。

Answer 2

因为 x y=1,(x 2) (y 1)=4

那么x^2/(x 2) y^2/(y 1)
=[(x 2)-2]^2/(x 2) [(y 1)-1]^2/(y 1)
=[(x 2)^2-4(x 2) 4]/(x 2) [(y 1)^2-2(y 1) 1]/(y 1)
=(x 2)-4 4/(x 2) (y 1)-2 1/(y 1)
=(x y-3) 4/(x 2) 1/(y 1)
=-2 [(x 2) (y 1)]/(x 2) [(x 2) (y 1)]/[4(y 1)]
=-2 1 1/4 (y 1)/(x 2) (x 2)/[4(y 1)]
=-3/4 (y 1)/(x 2) (x 2)/[4(y 1)]

∵(y 1)/(x 2) (x 2)/[4(y 1)]≥2√(1/4)=1
当且仅当(y 1)/(x 2)=(x 2)/[4(y 1)]
即x 2=2(y 1)
x=2/3,y=1/3时取等号

∴x^2/(x 2) y^2/(y 1)≥1/4
即最小值为1/4