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Fie numerele reale \( x\ ,\ y\ ,\ x_1\ ,\ y_1 \) pentru care \( 1<x<y \) si \( \left\{\begin{array}{c}
x_1=x+\frac yx-1\\\\
y_1=y+\frac xy-1\end{array} \) .

Sa se arate ca : \( \left\|\begin{array}{cc}
1. & \{\ x_1\ ,\ y_1\ \}\subset (\ x\ ,\ y\ )\\\\
2. & x_1=y_1\ \Longleftrightarrow\left\|\begin{array}{c}
x=y=2\\\\
\mathrm {\ sau\ }\\\\
2\in (\ x\ ,\ y\ )\end{array}\end{array} \) .

1. \( y>x \) rezulta \( \frac{y}{x}>1 \) rezulta \( x_1=x+\frac{y}{y}-1>x \)
\( (x-y)(x-1)<0 \) echivalent cu \( x(x-y)+y-x<0 \) echivalent cu \( x^2+y-x<xy \)
\( x+\frac{y}{x}-1<y \) adica \( x_1<y \)

deci \( x_1\in\(x,y) \)
Analog si \( y_1\in\(x,y) \)


2. \( x_1=y_1 \)
\( x+\frac{y}{x}-1=y+\frac{x}{y}-1 \)
\( x^2y+y^2=y^2x+x^2 \)
\( xy(x-y)=(x+y)(x-y) \)

I. \( x=y \) cu toate ca in enunt se dadea strict mai mare x ca y...
dar din \( x=y \) => \( x_1=y_1 \) oricare ar fi x real...

II. \( x\neq{y} \)
\( xy=x+y \)
\( (x-1)(y-1)=1 \) de unde rezulta \( 1\in\(x,y) \)