//Điểm rơi : x=2 và y=4
Ta có : \(A=3a+2b+\frac{6}{a}+\frac{8}{b}\)
\(=\left(\frac{3a}{2}+\frac{6}{a}\right)+\left(\frac{b}{2}+\frac{8}{b}\right)+\frac{3}{2}\left(a+b\right)\)
\(\ge2\sqrt{\frac{3a}{2}.\frac{6}{a}}+2\sqrt{\frac{b}{2}.\frac{8}{b}}+\frac{3}{2}.6\)
\(=2\sqrt{9}+2\sqrt{4}+9\)
\(=2.3+2.2+9=19\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\frac{3a}{2}=\frac{6}{a}\\\frac{b}{2}=\frac{8}{b}\\a+b=6\end{cases}}\Leftrightarrow\hept{\begin{cases}a=2\\b=4\end{cases}}\)
Vậy \(A_{min}=19\Leftrightarrow\hept{\begin{cases}a=2\\b=4\end{cases}}\)