OLM Neamt 1998
Posted: Sat Nov 08, 2008 11:35 pm
Daca \( a,b,c \in\mathbb{Q}_+^* \) si \( \frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b} \), calculati \( \frac{a^2+b^2+c^2}{ab+ac+bc}. \)
\( \frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{a+b+b+c+a+c}=\frac{a+b+c}{2(a+b+c)}=\frac{1}{2}\Rightarrow
a+b=2c; a+c=2b; b+c=2a. \)
De unde \( c=2a-b \) si \( c=2a-b \)\( \rightarrow2a-b=2b-a\Rightarrow 3a=3b\Rightarrow a=b \), adica \( ab=a\cdot a=a^2; bc=b\cdot b=b^2; ac=c\cdot c=c^2 \).
Atunci \( \frac{a^2+b^2+c^2}{ab+bc+ac}=\frac{a^2+b^2+c^2}{a^2+b^2+c^2}=1 \).