Daca \( a,b,c \in \mathbb{R}^* \) astfel incat \( (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=1 \) , sa se arate ca :
\( (a+b)(b+c)(c+a)=0 \) .
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Harghita 2006
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Harghita 2006
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Adriana Nistor
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Desfacem parantezele si obtinem:
\( \frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}+3=1 \),deci \( \frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{a}{c}+\frac{c}{a}+2=0 \);
\( \frac{a^2c*c+b^2*c+b^2*a+c^2*a+c^2*b+a^2*b+2abc}{abc}=0 \)
\( \frac{ab(a+b)+c^2(a+b)+c(a+b)^2}{abc}=0 \); \( \frac{(a+b)(c^2+ca+cb+ab)}{abc}=0 \),
adica \( \frac{(a+b)(b+c)(c+a)}{abc}=0 \),de unde \( (a+b)(b+c)(c+a)=0 \)
\( \frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}+3=1 \),deci \( \frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{a}{c}+\frac{c}{a}+2=0 \);
\( \frac{a^2c*c+b^2*c+b^2*a+c^2*a+c^2*b+a^2*b+2abc}{abc}=0 \)
\( \frac{ab(a+b)+c^2(a+b)+c(a+b)^2}{abc}=0 \); \( \frac{(a+b)(c^2+ca+cb+ab)}{abc}=0 \),
adica \( \frac{(a+b)(b+c)(c+a)}{abc}=0 \),de unde \( (a+b)(b+c)(c+a)=0 \)