Исбот кунед, ки \(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}>\frac{y}{x}+\frac{z}{y}+\frac{x}{z}\),

агар \(0\lt x\lt y\lt z\) бошад.

Азбаски \(0\lt x\lt y\lt z\), мешавад ки

\(\frac{x}{y}\lt1,\)

\(\frac{y}{z}\lt1,\)

\(\frac{z}{x}>1.\)

Бинобар ҳамин:

\(\frac{x}{y}-1\lt0,\)

\(\frac{y}{z}-1\lt0,\)

\(\frac{z}{x}-1>0\) ва

\((\frac{x}{y}-1)\cdot(\frac{y}{z}-1)\cdot(\frac{z}{x}-1)>0.\)

\((\frac{x}{y}-1)\cdot(\frac{y}{z}-1)\cdot(\frac{z}{x}-1)=(\frac{x}{y}-1)\cdot(\frac{y}{x}-\frac{y}{z}-\frac{z}{x}+1)=\)

\(=\frac{x}{y}\cdot\frac{y}{x}-\frac{x}{y}\cdot\frac{y}{z}-\frac{x}{y}\cdot\frac{z}{x}+\frac{x}{y}-\)

\(-\frac{y}{x}+\frac{y}{z}+\frac{z}{x}-1=\)

\(=1-\frac{x}{z}-\frac{z}{y}+\frac{x}{y}-\frac{y}{x}+\frac{y}{z}+\frac{z}{x}-1=\)

\(=\frac{x}{y}+\frac{y}{z}+\frac{z}{x}-(\frac{y}{x}+\frac{z}{y}+\frac{x}{z})\)

\(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}-(\frac{y}{x}+\frac{z}{y}+\frac{x}{z})>0\)

Яъне

\(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}>\frac{y}{x}+\frac{z}{y}+\frac{x}{z}\)

Исбот шуд.