Question

CMR:

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

Answer 1

Đặt:

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

Ta có:

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

Cộng 3 vế ta có:

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

Ta lại có:

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

Cộng 3 vế lại ta có:

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