Question

已知x>0,y>0,3/(2+x)+3/(2+y)=1,求x+3y的最小值

Answer 1

解： ∵3/(2+x)+3/(2+y)=1
∴y=(x+8)/(x-1)
∴x+3y=x+3(x+8)/(x-1)
=(x^2-x+3x+24)/(x-1)
=(x^2+2x+24)/(x-1)
=[(x-1)^2+4(x-1)+27]/(x-1)
=(x-1)+27/(x-1)+4
≥2√27+4=4+6根号3
当x=1+3根号3时，上式等号成立
所以最小值为4+6根号3

Answer 2

最小值是6√3＋4由3/(2 x) 3/(2 y)=1得y＝（8 x）/（x－1）令f（x）＝x＋3y将y带入并对f（x）求导，得出的求导式子使其等于零可得出最小值

Answer 3

3/(2+x)+3/(2+y)=1 , x>1,y>1
通分得：3(4+x+y)=(x+2)(y+2)
xy-x-y-12=0
y=(x+12)/(x-1)

x+3y=x+3(x+12)/(x-1) = [x(x-1)+3(x+12)]/(x-1)
= (x²+2x+36)/(x-1)
= (x²-2x+1+4x-4+39)/(x-1)
= (x-1) + 39/(x-1) + 4
≥ 2√39 + 4