Cho [tex]1\leq x< y\leq 2[/tex]. Tìm GTLN của A = [tex](x+y)(\frac{1}{x}+\frac{1}{y})[/tex]
Từ [tex]1\leq x< y\leq 2\Rightarrow \left\{\begin{matrix} 1<\frac{y}{x}\\ 2\geq \frac{y}{x} \end{matrix}\right. \Rightarrow 1< \frac{y}{x}\leq 2[/tex]
[tex]A=2+\frac{y}{x}+\frac{x}{y}[/tex]
Đặt [tex]\frac{y}{x}=t(1<t\leq 2)\Rightarrow A=2+t+\frac{1}{t}[/tex]
Xét [tex]A-\frac{9}{2}=\frac{t^2+2t+1}{t}-\frac{9}{2}=\frac{2t^2+4t+2-9t}{2t}=\frac{(2t-1)(t-2)}{2t}\leq \frac{3.0}{2t}=0\Rightarrow A_{max}=\frac{9}{2}[/tex]
Dấu = khi [tex]t=2\Leftrightarrow ...[/tex]