-Áp dụng BĐT AM-GM ta có:
\(2x.3y\le\dfrac{\left(2x+3y\right)^2}{4}\Rightarrow6xy\le\dfrac{\left(2x+3y\right)^2}{4}\le\dfrac{2^2}{4}=1\)
\(\Rightarrow xy\le\dfrac{1}{6}\)
\(A=\dfrac{4}{4x^2+9y^2}+\dfrac{9}{xy}=\dfrac{4}{4x^2+9y^2}+\dfrac{4}{12xy}+\dfrac{107}{12xy}=4\left(\dfrac{1}{4x^2+9y^2}+\dfrac{1}{12xy}\right)+\dfrac{26}{3xy}\)
-Áp dụng BĐT Caushy Schwarz ta có:
\(A=4\left(\dfrac{1}{4x^2+9y^2}+\dfrac{1}{12xy}\right)+\dfrac{26}{3xy}\ge4.\dfrac{\left(1+1\right)^2}{4x^2+9y^2+12xy}+\dfrac{26}{3xy}=4.\dfrac{4}{\left(2x+3y\right)^2}+\dfrac{26}{3xy}\ge4.\dfrac{4}{2^2}+\dfrac{26}{3.\dfrac{1}{6}}=56\)
\(A=56\Leftrightarrow\left\{{}\begin{matrix}2x=3y\\2x+3y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)
-Vậy \(A_{min}=56\)
