已知x>0y>0,x+2y+2xy=8则x+2y最小值为

问题描述:

已知x>0y>0,x+2y+2xy=8则x+2y最小值为

x+2y+2xy=8≥2√2xy+2xy,
令√2xy=a,∴a^2+2a≦8,得;
(a+4)(a-2)≦0,a>0,∴a≤2
x+2y≥2√2xy≥4