Question

Cho các số nguyên dương x;y;z

CM : \(1< \frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}< 2\)

Answer 1

Ta có: \(\left\{\begin{matrix}\frac{x}{x+y}>\frac{x}{x+y+z}\\\frac{y}{y+z}>\frac{y}{x+y+z}\\\frac{z}{z+x}>\frac{z}{x+y+z}\end{matrix}\right.\Rightarrow\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}>\frac{x}{x+y+z}+\frac{y}{x+y+z}+\frac{z}{x+y+z}=1\)

\(\left\{\begin{matrix}\frac{x}{x+y}< \frac{x+z}{x+y+z}\\\frac{y}{y+z}< \frac{x+y}{x+y+z}\\\frac{z}{z+x}< \frac{y+z}{x+y+z}\end{matrix}\right.\Rightarrow\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}< \frac{x+y}{x+y+z}+\frac{x+y}{x+y+z}+\frac{y+z}{x+y+2}=2\)

\(\Rightarrow1< \frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}< 2\left(đpcm\right)\)