vumin

 Không mất tính tổng quát,giả sử$\frac{1}{2}\geq x\geq y\geq 0$

 Ta thấy:$P(x,y)\leq \frac{1}{4}(\frac{x}{y+1}+\frac{y}{x+1})\leq \frac{1}{4} \frac{x+y}{y+1}\leq \frac{1}{6}(0\leq y\leq x\leq \frac{1}{2})$

 Vậy max P=1/6 khi x=y=0,5

  Ta có:$\underset{BD}{\rightarrow}=\frac{3}{5}underset{BC}{\rightarrow}\Rightarrow \underset{BE}{\rightarrow}+\underset{ED}{\rightarrow}=\frac{3}{5}\underset{BE}{\rightarrow}+\frac{3}{5}\underset{EC}{\rightarrow}\Rightarrow 5\underset{ED}{\rightarrow}=4\underset{EA}{\rightarrow}\Rightarrow A,D,E $ thẳng hàng(ĐPCM)