OLM Neamt 1998

[alex2008]

Daca \( a,b,c \in\mathbb{Q}_+^* \) si \( \frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b} \), calculati \( \frac{a^2+b^2+c^2}{ab+ac+bc}. \)

[Andi Brojbeanu]

\( \frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{a+b+b+c+a+c}=\frac{a+b+c}{2(a+b+c)}=\frac{1}{2}\Rightarrow
a+b=2c; a+c=2b; b+c=2a. \)
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 \( \frac{a^2+b^2+c^2}{ab+bc+ac}=\frac{a^2+b^2+c^2}{a^2+b^2+c^2}=1 \).

[Virgil Nicula]

\( \frac {a}{b+c}=\frac {b}{c+a}=\frac {c}{a+b}=\frac {a+b+c}{2(a+b+c)}=\frac 12=\frac {a^2}{a(b+c)}= \) \( \frac {b^2}{b(c+a)}=\frac {c^2}{c(a+b)}=\frac {a^2+b^2+c^2}{2(ab+bc+ca)}\ \Longrightarrow\ \frac {a^2+b^2+c^2}{ab+bc+ca}=1\ . \)