证明：∵a＞b＞0,且a²=a(a-b)+ab.∴由基本不等式得：a²+(1/ab)+[1/a(a-b)]=a(a-b)+ab+(1/ab)+[1/a(a-b)]≥4√{a(a-b)ab×1/ab×1/a(a-b)=4.等号仅当a=√2,b=√2/2时取得.∴a²+1/ab+1/a(a-b)≥4....