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Stiind ca \( x+2z+1>0 \) , \( y+2x+2>0 \) , \( z+2y+3>0 \) , sa se rezolve in\( \mathbb{R} \) sistemul :

\( \left{\begin{array}{cc}x+y+z=1\\\frac{x+2y+2}{x+2z+1}+\frac{y+2z+1}{y+2x+2}+\frac{z+2x}{z+2y+3}=2\end{array} \)

V Berghea,G.M.

Indicatie:

Noteaza \( x+2z+1=a \) , \( y+2x+2=b \) , \( z+2y+3=c \)

\( a+b+c=9 \)

Atunci prima ecuatie se scrie \( \frac{b+c-4}{a}+\frac{c+a-4}{b}+\frac{b+a-4}{c}=2 \)