Câu 5:
\(A=\dfrac{2}{x}+\dfrac{6}{y}+\dfrac{9}{3x+y}\)
\(=\dfrac{2\left(3x+y\right)}{xy}+\dfrac{9}{3x+y}\)
\(=\dfrac{3x+y}{6}+\dfrac{9}{3x+y}\left(xy=12\right)\)
\(=\dfrac{3x+y}{16}+\dfrac{9}{3x+y}+\dfrac{5\left(3x+y\right)}{48}\)
\(\ge2\sqrt{\dfrac{3x+y}{16}.\dfrac{9}{3x+y}}+\dfrac{5.2\sqrt{3x.y}}{48}\)
\(=2\sqrt{\dfrac{9}{16}}+\dfrac{10\sqrt{3.12}}{48}=\dfrac{11}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\end{matrix}\right.\)
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