请问老师：x,y是正实数,x+y=1,求x^2/x+2+y^2/y+1的最小请问老师：x,y是正实数,x+y=1,求x^2/x+2+y^2/y+1的最小值.（老师,我用了换元求导,虽然得出答案但很烦,请老师教我用基本不等式做）谢谢

Question

请问老师：x,y是正实数,x+y=1,求x^2/x+2 + y^2/y+1 的最小
请问老师：x,y是正实数,x+y=1,求x^2/x+2 + y^2/y+1 的最小值.
（老师,我用了换元求导,虽然得出答案但很烦,请老师教我用基本不等式做）谢谢

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