Tìm min

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locxoaymgk

Cho x,y,z>0 và x+y+z=1.
Tìm min [TEX]T=\sqrt{\frac{1-x}{1+x}}+\sqrt{\frac{1-y}{1+y}}+\sqrt{\frac{1-z}{1+z}}[/TEX]

Theo cô si ta có:

[TEX]2\sqrt{\frac{1+x}{1-x}}.1\leq\frac{1+x+1-x}{1-x}=\frac{2}{1-x}.[/TEX]

[TEX]\Rightarrow \sqrt{\frac{1+x}{1-x}} \leq \frac{1}{1-x}.[/TEX]

[TEX]\Rightarrow \sqrt{\frac{1-x}{1+x}}\geq1-x.[/TEX]

CMTT rồi cộng từng vế 3 BDT ta có:

[TEX] \sqrt{\frac{1-x}{1+x}}+\sqrt{\frac{1-y}{1+y}}+\sqrt{\frac{1-z}{1+z}} \geq 3-(x+y+z)=2.[/TEX]

Dấu = xảy ra \Leftrightarrow ...
 
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