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truongduong9083

Chào bạn

Ta có
+ [TEX]\frac{x+y}{1+z}+\frac{y+z}{1+x}+\frac{z+x}{1+y}= \frac{x}{1+z}+\frac{z}{1+x}+\frac{y}{1+z}+\frac{z}{1+y}+\frac{y}{1+x}+\frac{x}{1+y} \leq \frac{x}{x+z}+\frac{z}{z+x}+\frac{y}{y+z}+\frac{z}{z+y}+\frac{y}{y+x}+\frac{x}{x+y} = 3[/TEX]
+ Ta có [TEX]1+z \leq x+y+z; 1+y \leq x+y+z; 1+x \leq x+y+z[/TEX]
[TEX]\Rightarrow\frac{x+y}{1+z}+\frac{y+z}{1+x}+\frac{z+x}{1+y}\geq \frac{x+y}{x+y+z}+\frac{y+z}{z+y+x}+\frac{z+x}{z+x+y} = 2[/TEX]
 
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