设a＞b＞0，则a2+1ab+1a(a-b)的最小值是（　　）

a2+1ab+1a(a-b)=ab+1ab+a(a-b)+1a(a-b)≥4当且仅当ab=1aba(a-b)=1a(a-b)取等号即a=2b=22取等号．∴a2+1ab+1a(a-b)的最小值为4故选：D