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Fie \( a \), \( b \), \( c\in\mathbb{R}^{\ast} \) si \( x \), \( y \), \( z \), \( k \), \( l \), \( m \), \( n\in\mathbb{R} \).

Daca \( x^2+y^2+z^2+l^2+m^2+n^2+k^2(a^2+b^2+c^2)+2k(al+bm+cn)= \)

\( 2k(ax+by+cz)+2(lx+my+nz) \), atunci \( \frac{x-l}a=\frac{y-m}b=\frac{z-n}c=k. \)

Proportia ce trebuie demonstrata ma duce cu gandul la conditia de egalitate din inegalitatea C-B-S.

\( x^2+y^2+z^2+l^2+m^2+n^2+k^2(a^2+b^2+c^2)+2k(al+bm+cn)=x^2+y^2+z^2+(l^2+k^2a^2+2kal)+(m^2+k^2b^2+2kbm)+(n^2+k^2c^2+2kcn)= \)
\( =[x^2+(l+ka)^2]+[y^2+(m+kb)^2]+[z^2+(n+kc)^2]\geq 2x(l+ka)+2y(m+kb)+2z(n+kc)=2k(ax+by+cz)+2(xl+ym+zn) \).
Conform ipotezei, avem egalitate:
\( x^2+(l+ka)^2=2x(l+ka)\Rightarrow x^2+(l+ka)^2-2x(l+ka)=0; [x-(l+ka)]^2=0\Rightarrow x=l+ka; x-l=ka\Rightarrow \frac{x-l}{a}=k^{(1)} \).
\( y^2+(m+kb)^2=2y(m+kb)\Rightarrow y^2+(m+kb)^2-2y(m+kb)=0; [y-(m+kb)]^2=0\Rightarrow y=m+kb; y-m=kb\Rightarrow \frac{y-m}{b}=k^{(2)} \).
\( z^2+(n+kc)^2=2z(n+kc)\Rightarrow z^2+(n+kc)^2-2z(n+kc)=0; [z-(n+kc)]^2=0\Rightarrow z=n+kc; z-n=kc\Rightarrow \frac{z-n}{c}=k^{(3)} \).
Din relatiile \( (1),(2),(3) \) obtinem ca \( \frac{x-l}{a}=\frac{y-m}{b}=\frac{z-n}{c}=k \).