Question

设对任意实数x>0,y>0。若不等式x+√xy≤a(x+2y)恒成立,则实数a的最小值为

Answer 1

√(xy)≤(a-1)x+2ay
xy≤(a-1)^2x^2+4a^2y^2+4a(a-1)xy
(a-1)^2x^2+4a^2y^2+[4a(a-1)-1]xy≥0
4a^2(y/x)^2+(4a^2-4a-1)(y/x)+
(a-1)^2
≥0
若上式恒成立需满足:
(-4a^2+4a+1)/(8a^2)≤0
解得a≥(1+√2)/2
故a的最小值为(1+√2)/2。