\(A=\frac{4}{4x^2+9y^2}+\frac{9}{xy}=\frac{4}{4x^2+9y^2}+\frac{54}{6xy}\)
Đặt \(\left\{{}\begin{matrix}2x=a\\3y=b\end{matrix}\right.\Rightarrow A=\frac{4}{a^2+b^2}+\frac{54}{ab}\)
\(A=\frac{4}{a^2+b^2}+\frac{4}{2ab}+\frac{52}{ab}\)
\(A=4\left(\frac{1}{a^2+b^2}+\frac{1}{2ab}\right)+\frac{52}{ab}\)
\(\ge\frac{16}{\left(a+b\right)^2}+\frac{52}{\frac{\left(a+b\right)^2}{4}}\ge4+52=56\)
\("="\Leftrightarrow a=b\Leftrightarrow2x=3y\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\y=\frac{1}{3}\end{matrix}\right.\)
\(A=\frac{4}{4x^2+9y^2}+\frac{4}{12xy}+\frac{52}{2x.3y}\ge\frac{16}{4x^2+9y^2+12xy}+\frac{52}{\frac{\left(2x+3y\right)^2}{4}}\)
\(A\ge\frac{16}{\left(2x+3y\right)^2}+\frac{208}{\left(2x+3y\right)^2}\ge\frac{16}{4}+\frac{208}{4}=56\)
\(\Rightarrow A_{min}=56\) khi \(\left\{{}\begin{matrix}x=\frac{1}{2}\\y=\frac{1}{3}\end{matrix}\right.\)
