Áp dụng BĐT Cauchy
\(1\ge a+b\ge2\sqrt{ab}\Leftrightarrow1\ge4ab\Leftrightarrow ab\le\dfrac{1}{4}\)
Ta có:
\(A=ab+\dfrac{1}{ab}=ab+\dfrac{1}{16ab}+\dfrac{15}{16ab}\)
\(\ge2.\sqrt{ab.\dfrac{1}{16ab}}+\dfrac{15}{16.\dfrac{1}{4}}=\dfrac{17}{4}\)
Vậy \(A_{MIN}=\dfrac{17}{4}\) khi và chỉ khi \(a=b=\dfrac{1}{2}\)