Question

若实数x,y满足:x,y>0,3xy-x-y-1%=0,则xy的最小值为多少？

Answer 1

x,y>0,3xy-x-y-1%=0,
所以3xy-1%=x+y≥2√(xy),
设u=√(xy),上式化为3u^2-2u-1%≥0,
解得u≥(1+√1.03)/3
当x=y=(1+√1.03)/3时取等号，
所以xy的最小值=[(1+√1.03)/3]^2=(2.03+2√1.03)/9.

Answer 2

3xy-x-y-1=0
y(3x-1)=x+1
y=(x+1)/(3x-1)
所以xy=(x^2+x)/(3x-1)
=[(x-1/3)^2+(5/3)*(x-1/3)+4/9]/(3x-1)
=(1/3)*(x-1/3)+(4/27)*1/(x-1/3)+5/9
>=2√[(1/3)*(4/27)]+5/9
=1
当且仅当(1/3)*(x-1/3)=(4/27)*1/(x-1/3)，即x=1，y=1时，取到最小值1