Question

已知x,y为正实数,且4/（x+1）+1/（2y+1）=1,则x+2y的最小值为

Answer 1

x+2y

=(x+1)+(2y+1)-2
=[(x+1)+(2y+1)][4/(x+1)+1/(2y+1)]-2
=4+(x+1)/(2y+1)+4(2y+1)/(x+1)+1-2
>=2根号[(x+1)/(2y+1)*4(2y+1)/(x+1)]+3
=2*2+3
=7
即最小值是:7

Answer 2

4/(x+1)+1/(2y+1)=1
1/(2y+1) = (x-3)/(x+1)
2y+1= (x+1)/(x-3)
y = 2/(x-3)
S=x+2y
=x + 4/(x-3)
S' = 1-4/(x-3)^2 =0
x^2-6x+5=0
x=1 or 5
S''=8/(x-3)^3
S''(5)>0 (mn)
x=5, y=1
minS = 5+2 =7