Question

cm:

\(1< \dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}< 2\)

Answer 1

Đặt:

\(linh=\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}\)

Ta có: \(\left\{{}\begin{matrix}\dfrac{x}{x+y}>\dfrac{x}{x+y+z}\\\dfrac{y}{y+z}>\dfrac{y}{x+y+z}\\\dfrac{z}{x+z}>\dfrac{z}{x+y+z}\end{matrix}\right.\)

Cộng theo 3 vế ta có:

\(linh>\dfrac{x}{x+y+z}+\dfrac{y}{x+y+z}+\dfrac{z}{x+y+z}=1\)(1)

Lại có: \(\left\{{}\begin{matrix}\dfrac{x}{x+y}< \dfrac{x+z}{x+y+z}\\\dfrac{y}{y+z}< \dfrac{x+y}{x+y+z}\\\dfrac{z}{x+z}< \dfrac{y+z}{x+y+z}\end{matrix}\right.\)

Cộng theo 3 vế ta có:
\(linh< \dfrac{x+z}{x+y+z}+\dfrac{x+y}{x+y+z}+\dfrac{y+z}{x+y+z}=2\) (2)
Từ (1) và (2) ta có:

\(1< linh< 2\left(đpcm\right)\)