G
giaosu_fanting_thientai
[TEX]Dieu\ kien\ :\ \left{ x \geq 1 \\ y \geq 1[/TEX]
[TEX](pt) \Leftrightarrow \frac{\sqrt{y-1}}{y}+\frac{2\sqrt{x-1}}{x}=\frac{3}{2}[/TEX]
Áp dụng BDT Co-si ta có:
[TEX]\frac{\sqrt{y-1}}{y} \leq \frac{(y-1)+1}{2y} =\frac{1}{2}[/TEX]
[TEX]\frac{2\sqrt{x-1}}{x} \leq \frac{(x-1)+1}{x} =1[/TEX]
[TEX]\Rightarrow \Leftrightarrow \frac{\sqrt{y-1}}{y}+\frac{2\sqrt{x-1}}{x} \leq \frac{3}{2}[/TEX]
[TEX]Dau\ "="\ \Leftrightarrow \left{ x=2\\ y=2[/TEX]
[TEX](pt) \Leftrightarrow \frac{\sqrt{y-1}}{y}+\frac{2\sqrt{x-1}}{x}=\frac{3}{2}[/TEX]
Áp dụng BDT Co-si ta có:
[TEX]\frac{\sqrt{y-1}}{y} \leq \frac{(y-1)+1}{2y} =\frac{1}{2}[/TEX]
[TEX]\frac{2\sqrt{x-1}}{x} \leq \frac{(x-1)+1}{x} =1[/TEX]
[TEX]\Rightarrow \Leftrightarrow \frac{\sqrt{y-1}}{y}+\frac{2\sqrt{x-1}}{x} \leq \frac{3}{2}[/TEX]
[TEX]Dau\ "="\ \Leftrightarrow \left{ x=2\\ y=2[/TEX]
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