Daca \( a, b, c \in (0;1) \) sau \( a, b, c \in (1; \infty) \), atunci aratati ca: \( \log_{a^2b}a+\log_{b^2c}b+\log_{c^2a}c\leq 1 \).
OLM Constanta 2008, Gazeta Matematica
(Pentru \( \sum log_{ab^2} \) inegalitatea devine \( \geq 1 \).)
Inegalitate cu logaritmi
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Andrei Velicu
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Radu Titiu
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Notez x=ln(a), etc atunci inegalitatea de demonstrat devine:
\( \sum \frac{x}{2x+y}\leq 1 \Leftrightarrow \sum \frac{1}{2}\left( 1-\frac{y}{2x+y}\right) \leq 1 \) \( \Leftrightarrow \sum \frac{y}{2x+y}\geq 1 \).
Dar din inegalitatea CBS avem:
\( \sum \frac{y}{2x+y}=\sum \frac{y^2}{2xy+y^2} \geq \frac{(x+y+z)^2}{x^2+y^2+z^2+2(xy+yz+xz)}=1 \).
Procedand analog se poate deomnstra urmatoarea generalizare:
Daca \( k\geq 2p>0 \) si \( a,b,c \in (1,\infty) \) atunci \( \sum log_{a^kb^p}a\leq \frac{3}{k+p} \).
\( \sum \frac{x}{2x+y}\leq 1 \Leftrightarrow \sum \frac{1}{2}\left( 1-\frac{y}{2x+y}\right) \leq 1 \) \( \Leftrightarrow \sum \frac{y}{2x+y}\geq 1 \).
Dar din inegalitatea CBS avem:
\( \sum \frac{y}{2x+y}=\sum \frac{y^2}{2xy+y^2} \geq \frac{(x+y+z)^2}{x^2+y^2+z^2+2(xy+yz+xz)}=1 \).
Procedand analog se poate deomnstra urmatoarea generalizare:
Daca \( k\geq 2p>0 \) si \( a,b,c \in (1,\infty) \) atunci \( \sum log_{a^kb^p}a\leq \frac{3}{k+p} \).
A mathematician is a machine for turning coffee into theorems.