OLM 2008 ARGES

Moderators: Filip Chindea, Andrei Velicu, Radu Titiu

Post Reply
Adriana Nistor
Pitagora
Posts: 82
Joined: Thu Aug 07, 2008 10:07 pm
Location: Drobeta Turnu Severin, Mehedinti

OLM 2008 ARGES

Post by Adriana Nistor »

Daca \( a,b,c >1 \) sau \( a,b,c\in (0,1) \) si \( n\ge 2, n\in N \), sa se arate ca:
a) \( \log_{ab^n}{a} + \log_{bc^n}{b} + \log_{ca^n}{c} \ge \frac{3}{n+1} \);
b) \( \log_{a^nb}{a} + \log_{b^nc}{b} + \log_{c^na}{c}\le\frac{3}{n+1} \).
Theodor Munteanu
Pitagora
Posts: 98
Joined: Tue May 06, 2008 5:46 pm
Location: Sighetu Marmatiei

Post by Theodor Munteanu »

\(
\begin{array}{l}
\ln a = x,\ln b = y,\ln c = z \\ a) \Leftrightarrow \frac{x}{{x + ny}} + \frac{y}{{y + nz}} + \frac{z}{{z + nx}} \ge \frac{3}{{n + 1}}; \\
LHS = \frac{{x^2 }}{{x^2 + nxy}} + \frac{{y^2 }}{{y^2 + nyz}} + \frac{{z^2 }}{{z^2 + nxz}} \ge \frac{{(x + y + z)^2 }}{{x^2 + y^2 + z^2 + n(xy + yz + zx)}}; \\
\frac{{(x + y + z)^2 }}{{x^2 + y^2 + z^2 + n(xy + yz + zx)}} \ge \frac{3}{{n + 1}}(*);(*) \Leftrightarrow (n + 1)\left( {x^2 + y^2 + z^2 } \right) + (2n + 2)(xy + yz + zx) \ge 3\left( {x^2 + y^2 + z^2 } \right) + 3n(xy + yz + zx); \\
b) \frac{x}{{nx + y}} + \frac{y}{{ny + z}} + \frac{z}{{nz + y}} \le \frac{3}{{n + 1}}(**); \\
LHS = \frac{1}{n}\left( {1 - \frac{y}{{nx + y}}} \right) + \frac{1}{n}\left( {1 - \frac{z}{{ny + z}}} \right) + \frac{1}{n}\left( {1 - \frac{x}{{nz + x}}} \right) \Rightarrow (**) \Leftrightarrow 3 - \left( {\frac{y}{{nx + y}} + \frac{z}{{ny + z}} + \frac{x}{{nz + x}}} \right) \le \frac{{3n}}{{n + 1}} \Leftrightarrow \frac{y}{{nx + y}} + \frac{z}{{ny + z}} + \frac{x}{{nz + x}} \ge \frac{3}{{n + 1}}(\alpha ); \\
LHS = \frac{{y^2 }}{{nxy + y^2 }} + \frac{{z^2 }}{{nyz + z^2 }} + \frac{{x^2 }}{{nzx + x^2 }} \ge \frac{{(x + y + z)^2 }}{{n(xy + yz + zx) + x^2 + y^2 + z^2 }} \ge \frac{3}{{n + 1}} \\
\end{array}
\)
La inceput a fost numarul. El este stapanul universului.
Post Reply

Return to “Clasa a X-a”