[tex]P=\frac{3x}{y(x+1)}+\frac{3y}{x(y+1)}-\frac{1}{x^2}-\frac{1}{y^2}=\frac{3xy}{y^2(y+1)}+\frac{3xy}{x^2(y+1)}-\frac{1}{x^2}-\frac{1}{y^2}=\frac{x+y+1}{y^2(x+1)}+\frac{x+y+1}{x^2(y+1)}=\frac{1}{y(x+1)}+\frac{1}{x(y+1)}[/tex]
từ giả thiết ta có:
[tex]\frac{x+y+1}{xy}=3\Leftrightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{xy}=3[/tex]
đặt: [tex]\left\{\begin{matrix} \frac{1}{x}=a>0 & \\ \frac{1}{y}=b>0& \end{matrix}\right.[/tex]
=>[tex]a+b+ab=3 \Leftrightarrow (a+1)(b+1)=4[/tex]
=> [tex]P=\frac{ab}{a+1}+\frac{ab}{b+1}=\frac{ab}{4}(a+b+2)=\frac{3-a-b}{4}(a+b+2)[/tex]
đặt [tex]a+b=t>0[/tex]
=> [tex]P=\frac{(3-t)(t+2)}{4}=\frac{-t^2+t+6}{4}=\frac{\left ( -t^2+t-\frac{1}{4} \right )+\frac{25}{4}}{4}=\frac{-\left ( t-\frac{1}{2} \right )^2+\frac{25}{4}}{4}\leq \frac{25}{16}[/tex]
dấu "=" khi [tex]t=\frac{1}{2}\Rightarrow a+b=\frac{1}{2}[/tex]
giải hệ phương trình:
[tex]\left\{\begin{matrix} a+b=\frac{1}{2} & \\ a+b+ab=3 \Rightarrow ab=\frac{5}{2} & \end{matrix}\right.[/tex]
=> đến đây chắc rồi dùng vi-et ngược => a,b => x,y