设x>0,y>0,不等式1/x+1/y+m/(x+y)>=恒成立，则m的最小值是

1/x+1/y+m/(x+y)≥0
两边同时乘以x+y
(x+y)/x+(x+y)/y+m≥0
1+(y/x)+1+(x/y)+m≥0
(y/x)+(x/y)≥-(m+2)恒成立
因为(y/x)+(x/y)≥2√[(y/x)(x/y)]=2
所以2≥-(m+2)恒成立
2+m+2≥0
m≥-4
m的最小值是-4，当y/x=x/y，x=y时取最小值

因为x>0,y>0,所以xy>0
1/x+1/y+m/(x+y)=(x+y)/xy+m/(x+y)
>=2根号{[(x+y)/xy}{m/(x+y)]}
=2根号(m/xy)
若恒大于0,由m>=0

>=多少啊