Question

cho x,y dương thỏa mãn : \(x^2y+x+1\le y\) . Tìm Max của P=\(\dfrac{xy}{\left(x+y\right)^2}\)

Answer 1

từ giả thiết: \(x^2y+x+1\le y\Leftrightarrow y-x^2y-x-1\ge0\)

\(\Leftrightarrow y\left(1-x\right)\left(1+x\right)-\left(1+x\right)\ge0\)

\(\left(1+x\right)\left(y-xy-1\right)\ge0\)

x>0 ,\(\Rightarrow y-xy-1\ge0\Leftrightarrow1-x-\dfrac{1}{y}\ge0\)

\(1\ge x+\dfrac{1}{y}\ge2\sqrt{\dfrac{x}{y}}\Leftrightarrow\dfrac{x}{y}\le\dfrac{1}{4}\)

\(P=\dfrac{xy}{\left(x+y\right)^2}=\dfrac{1}{\dfrac{x}{y}+\dfrac{y}{x}+2}=\dfrac{1}{\dfrac{y}{x}+\dfrac{16x}{y}-\dfrac{15x}{y}+2}\le\dfrac{1}{10-\dfrac{15x}{y}}\le\dfrac{1}{10-\dfrac{15}{4}}=\dfrac{4}{25}\)dấu = xảy ra khi \(x=\dfrac{1}{y}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{2};y=2\)