$\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{4\sqrt{ab}}{a+b}
\\=\dfrac{a^2+b^2}{ab}+\dfrac{4\sqrt{ab}}{a+b}+\dfrac{4\sqrt{ab}}{a+b}-\dfrac{4\sqrt{ab}}{a+b}
\\\geq \dfrac{(a+b)^2}{2ab}+\dfrac{4\sqrt{ab}}{a+b}+\dfrac{4\sqrt{ab}}{a+b}-\dfrac{2(a+b)}{a+b}
\\\geq 3\sqrt[3]{ \dfrac{(a+b)^2}{2ab}.\dfrac{4\sqrt{ab}}{a+b}.\dfrac{4\sqrt{ab}}{a+b}}-2
\\=4$
Dấu '=' khi $a=b$