Question

cho x,y là 2 số dương thay đổi,thõa mãn xy=12.tìm min P= \(\frac{2}{x}+\frac{6}{y}+\frac{9}{3x+y}\)

Answer 1

\(xy=12\Rightarrow y=\frac{12}{x}\)

\(\Rightarrow P=\frac{2}{x}+\frac{x}{2}+\frac{3}{x+\frac{4}{x}}=\frac{1}{2}\left(x+\frac{4}{x}\right)+\frac{3}{x+\frac{4}{x}}\)

\(P=\frac{3}{16}\left(x+\frac{4}{x}\right)+\frac{3}{x+\frac{4}{x}}+\frac{29}{16}\left(x+\frac{4}{x}\right)\)

\(P\ge2\sqrt{\frac{9\left(x+\frac{4}{x}\right)}{16\left(x+\frac{4}{x}\right)}}+\frac{19}{16}.2\sqrt{x.\frac{4}{x}}=\frac{25}{4}\)

\(P_{min}=\frac{25}{4}\) khi \(\left\{{}\begin{matrix}x=2\\y=6\end{matrix}\right.\)