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

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

∵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