$A=(1+\frac{y}{x} )(\frac{x^2}{y^2}+1 )$Đặt $t=\frac{y}{x} \Rightarrow t\in [\frac{2011}{2012}, \frac{2012}{2011} ]$$\Rightarrow A=(1+t)(\frac{1}{t^2}+1 )$ $=\frac{1}{t^2} +1+\frac{1}{t} +t$Xét hàm $f(t)=\frac{1}{t^2} +\frac{1}{t}+t+1 $ $\Rightarrow f'(t)=-\frac{2}{t^3}-\frac{1}{t^2}+1 $ $f''(t)=\frac{6}{t^4}+\frac{2}{t^3} >0$$\Rightarrow f'(t)\leq f'(\frac{2012}{2011} )<0$$\Rightarrow f(t)$ nghịch biến$f(t)\geq f(\frac{2012}{2011} )$$f(t)_{min}=\frac{2012}{2011} $ $\Leftrightarrow \begin{cases}x=2011\\y=2012\end{cases} $
$A=(1+\frac{y}{x} )(\frac{x^2}{y^2}+1 )$Đặt $t=\frac{y}{x} \Rightarrow A\in [\frac{2011}{2012}, \frac{2012}{2011} ]$$\Rightarrow A=(1+t)(\frac{1}{t^2}+1 )$ $=\frac{1}{t^2} +1+\frac{1}{t} +t$Xét hàm $f(t)=\frac{1}{t^2} +\frac{1}{t}+t+1 $ $\Rightarrow f'(t)=-\frac{2}{t^3}-\frac{1}{t^2}+1 $ $f''(t)=\frac{6}{t^4}+\frac{2}{t^3} >0$$\Rightarrow f'(t)\leq f'(\frac{2012}{2011} )<0$$\Rightarrow f(t)$ nghịch biến$f(t)\geq f(\frac{2012}{2011} )$$f(t)_{min}=\frac{2012}{2011} $ $\Leftrightarrow \begin{cases}x=2011\\y=2012\end{cases} $
$A=(1+\frac{y}{x} )(\frac{x^2}{y^2}+1 )$Đặt $t=\frac{y}{x} \Rightarrow
t\in [\frac{2011}{2012}, \frac{2012}{2011} ]$$\Rightarrow A=(1+t)(\frac{1}{t^2}+1 )$ $=\frac{1}{t^2} +1+\frac{1}{t} +t$Xét hàm $f(t)=\frac{1}{t^2} +\frac{1}{t}+t+1 $ $\Rightarrow f'(t)=-\frac{2}{t^3}-\frac{1}{t^2}+1 $ $f''(t)=\frac{6}{t^4}+\frac{2}{t^3} >0$$\Rightarrow f'(t)\leq f'(\frac{2012}{2011} )<0$$\Rightarrow f(t)$ nghịch biến$f(t)\geq f(\frac{2012}{2011} )$$f(t)_{min}=\frac{2012}{2011} $ $\Leftrightarrow \begin{cases}x=2011\\y=2012\end{cases} $