Se considera numerele \( x,y,z \) pentru care \( xyz\ne 0 \) si \( \frac 1x+\frac 1y+\frac 1z=\frac {1}{xyz} \) .
Sa se dea un exemplu de numere \( x,y,z \) care verifica relatiile date. Sa se arate
ca sirurile \( 1+x^2\ ,\ 1+y^2\ ,\ 1+z^2 \) si \( y+z\ ,\ z+x\ ,\ x+y \) sunt invers proportionale.
Invers proportionalitate.
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Andi Brojbeanu
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\( x=1; y=\frac{1}{2}; z=\frac{1}{3} \).
\( \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{xyz}\Rightarrow xy+yz+zx=1. \)
\( (1+x^2)(y+z)=y+z+x(xy+xz)=y+z+x(1-yz)=y+z+x-xyz. \)
\( (1+y^2)(z+x)=z+x+y(yz+yx)=z+x+y(1-zx)=z+x+y-zxy. \)
\( (1+z^2)(x+y)=x+y+z(zx+zy)=x+y+z(1-xy)=x+y+z-zyx. \)
Atunci, \( (1+x^2)(y+z)=(1+y^2)(z+x)=(1+z^2)(x+y)=x+y+z-xyz\Rightarrow \)sirurile\( 1+x^2,\ 1+y^2,\ 1+z^2 \) si \( y+z,\ z+x,\ x+y \) sunt invers proportionale.
\( \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{xyz}\Rightarrow xy+yz+zx=1. \)
\( (1+x^2)(y+z)=y+z+x(xy+xz)=y+z+x(1-yz)=y+z+x-xyz. \)
\( (1+y^2)(z+x)=z+x+y(yz+yx)=z+x+y(1-zx)=z+x+y-zxy. \)
\( (1+z^2)(x+y)=x+y+z(zx+zy)=x+y+z(1-xy)=x+y+z-zyx. \)
Atunci, \( (1+x^2)(y+z)=(1+y^2)(z+x)=(1+z^2)(x+y)=x+y+z-xyz\Rightarrow \)sirurile\( 1+x^2,\ 1+y^2,\ 1+z^2 \) si \( y+z,\ z+x,\ x+y \) sunt invers proportionale.