已知正实数x,y满足x+2y=2.则y/2x+1/y的最小值是

问题描述:

已知正实数x,y满足x+2y=2.则y/2x+1/y的最小值是

由x+2y=2 得x/2+y=1
故y/2x+1/y=y/2x +(x/2+y)/y
=y/2x +x/2y+y/y
=y/2x +x/2y+1
由正数x,y
所以y/2x +x/2y+1>=1+2倍根号下(y/2x*x/2y)
=1+1
则y/2x+1/y>=2