Đặt \(D=\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}\)
Ta có:
\(\dfrac{x}{x+y}>\dfrac{x}{x+y+z}\)
\(\dfrac{y}{y+z}>\dfrac{y}{z+y+z}\)
\(\dfrac{z}{z+x}>\dfrac{z}{x+y+z}\)
\(\Rightarrow D>\dfrac{x}{x+y+z}+\dfrac{y}{x+y+z}+\dfrac{z}{x+y+z}=\dfrac{x+y+z}{x+y+z}=1\left(1\right)\)
\(\Rightarrow D>1\)
Ta lại có:
\(\dfrac{x}{x+y}< \dfrac{x+z}{x+y+z}\)
\(\dfrac{y}{y+z}< \dfrac{y+x}{x+y+z}\)
\(\dfrac{z}{z+x}< \dfrac{z+y}{x+y+z}\)
\(\Rightarrow D< \dfrac{x+y}{x+y+z}+\dfrac{y+x}{x+y+z}+\dfrac{x+y}{x+y+z}=\dfrac{x+z+y+x+z+y}{x+y+z}=\dfrac{2\left(x+y+z\right)}{z+y+z}=2\left(2\right)\) Từ (1) và (2) suy ra:
1 < D < 2
Suy ra : Biểu thức D không phải là số nguyên
