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

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

Answer 2

∵x+y=1
∴(x+2)+(y+1)=4

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)
=4/(x+2)+1/(y+1)+(x+y-3)
=[4/(x+2)+1/(y+1)] *[(x+2)+(y+1)]/4 -2
=[4+1+(x+2)/(y+1)+4(y+1)/(x+2)]/4-2
≥5+2√4-2=7
∴
x^2/(x+2)+y^2/(y+1)的最小值为7