Phải là x + y =6 nhé bn, x + y = 4 ko xảy ra dấu =
\(\Rightarrow2A=6x+4y+\frac{12}{x}+\frac{16}{y}\)
\(=3\left(x+y\right)+\left(3x+\frac{12}{x}\right)+\left(y+\frac{16}{y}\right)\)
AD BDT Cô-si cho 2 số không âm
\(\Rightarrow3x+\frac{12}{x}\ge2\sqrt{36}=12;y+\frac{16}{y}\ge2\sqrt{16}=8\)
\(\Rightarrow2A\ge3.6+12+8=38\)
\(\Rightarrow A\ge19\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)
Vậy \(A_{min}=19\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)