a>b>0,则a^2+1/ab+1/(a-b)的最小值是

a>b>0,则a^2+1/ab+1/(a-b)的最小值是
a>b>0,则a^2+1/ab+1/(a-b)的最小值是

a>b>0,则a^2+1/ab+1/(a-b)的最小值是
因a>b>0.故a²＞ab＞0.
===>a²-ab＞0,且ab＞0.
由基本不等式可知；
a²+(1/ab)+[1/(a²-ab)]
={(a²-ab)+[1/(a²-ab)]}+[(ab)+1/(ab)]≥2+2=4.
等号仅当a²-ab=1,ab=1时取得；
即当a=√2,b=1/√2时取得.故原式min=4.

a^2+1/ab+1/(a-b)=(a^2-b^2)+1/ab+1/(a-b)+b^2
=1/2(a-b)^2++1/(a-b)+1/2[(a^2+b^2)+1/ab)]+(1/2ab+2ab)
>=1/2[(a-b)^2++2/(a-b)]+1/2[(a^2+b^2)+2/(a^2+b^2)]+1/2（ab+4ab)
>=√2+√2+2