已知a>b>0,求a^2+1/(ab)+1/a(a-b)的最小值

Question

已知a>b>0,求a^2+1/(ab)+1/a(a-b)的最小值

Answer 1

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.