ĐK phải có thêm x,y>0 nữa chứ nhỉ
\(E=\frac{2013}{x}+\frac{1}{2013y}=\left(\frac{2013}{x}+2013x\right)+\left(\frac{1}{2013y}+2013y\right)-2013\left(x+y\right)\)
\(=\left(\frac{2013}{x}+2013x\right)+\left(\frac{1}{2013y}+2013y\right)-2013\cdot\frac{2014}{2013}\)
\(=\left(\frac{2013}{x}+2013x\right)+\left(\frac{1}{2013y}+2013y\right)-2014\)
Áp dụng bđt cô si ta có:
\(\frac{2013}{x}+2013x\ge2\sqrt{\frac{2013}{x}\cdot2013x}=2\cdot2013=4026\)
\(\frac{1}{2013y}+2013y\ge2\sqrt{\frac{1}{2013y}\cdot2013y}=2\)
Suy ra \(E\ge4026+2-2014=2014\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}\frac{2013}{x}=2013x\\\frac{1}{2013y}=2013y\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=\frac{1}{2013}\end{cases}}\)
Vậy...
Ta có
\(x+y=\frac{2014}{2013}\Rightarrow2013=\frac{2014}{x+y}\)
\(\Rightarrow E=\frac{2013}{x}+\frac{1}{2013y}=\frac{2014}{x\left(x+y\right)}+\frac{x+y}{2014y}\)
Áp dụng bđt AM-GM ta có
\(E\ge2\sqrt{\frac{2014\left(x+y\right)}{2014xy\left(x+y\right)}}=2\sqrt{\frac{1}{xy}}\)
Mặt khác \(xy\le\frac{\left(x+y\right)^2}{4}=\frac{2014^2}{\left(2.2013\right)^2}\)
\(\Rightarrow E\ge\sqrt{\frac{\left(2.2013\right)^2}{2014^2}}=\frac{4026}{2014}\)
Dấu "=" xảy ra khi\(x=y=\frac{2014}{4026}\)
