已知x,y属于R+,且2x+8y-xy=0,求x+y的最小值.
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已知x,y属于R+,且2x+8y-xy=0,求x+y的最小值.
2.已知x,y属于R+,且x+2y=3,求[1/(x+2)]+[1/2(y+1)]的最小值
2.已知x,y属于R+,且x+2y=3,求[1/(x+2)]+[1/2(y+1)]的最小值
利用重要不等式的性质
x,y>0,2x+8y=xy则2/y + 8/x =1则x+y=(x+y)(2/y + 8/x )
=2x/y +8y/x +10
> =8+10=18(均值不等式)
(当2x/y=8y/x即x=12,y=6时取=)
2)
x+2y=3
(x+2)+2y+2=7
[1/(x+2)]+[1/2(y+1)]
=[1/(x+2)]+[1/2(y+1)]*[(x+2)+2y+2]/7
=1/7 [1+2(y+1)/(x+2)+(x+2)/2(y+1)+1]
=1/7[2+2(y+1)/(x+2)+(x+2)/2(y+1)]
>=1/7[2+2根号1]
=4/7
当取得=时
2(y+1)=x+2
x,y>0,2x+8y=xy则2/y + 8/x =1则x+y=(x+y)(2/y + 8/x )
=2x/y +8y/x +10
> =8+10=18(均值不等式)
(当2x/y=8y/x即x=12,y=6时取=)
2)
x+2y=3
(x+2)+2y+2=7
[1/(x+2)]+[1/2(y+1)]
=[1/(x+2)]+[1/2(y+1)]*[(x+2)+2y+2]/7
=1/7 [1+2(y+1)/(x+2)+(x+2)/2(y+1)+1]
=1/7[2+2(y+1)/(x+2)+(x+2)/2(y+1)]
>=1/7[2+2根号1]
=4/7
当取得=时
2(y+1)=x+2
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