Fie numerele reale \( a, b, c \) astfel incat \( a, b, c>1 \) sau \( a, b, c\in (0,1) \). Sa se arate ca
\( \log_{a}bc+\log_{b}ca+\log_{c}ab\geq 4\left(\log_{ab}c+\log_{bc}a+\log_{ca}\right). \)
etapa judeteana, 2007, problema 1
Inegalitate simetrica cu logaritmi
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Inegalitate simetrica cu logaritmi
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.
Daca notam \( \log a=x ,\log b=y ,\log c=z \) (unde \( log \) are baza \( d \) care se afla in acelasi interval cu \( a,b,c \) astfel incat \( x,y,z \) sa fie pozitive) inegalitatea se rescrie sub forma echivalenta:
\( \displaystyle\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z} \geq 4 \left( \frac{z}{x+y}+\frac{y}{z+x}+\frac{x}{y+z} \right) \).
Folosind inegalitatea \( \frac {1}{x}+\frac{1}{y}\geq \frac{4}{x+y} \) rezulta concluzia.
\( \displaystyle\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z} \geq 4 \left( \frac{z}{x+y}+\frac{y}{z+x}+\frac{x}{y+z} \right) \).
Folosind inegalitatea \( \frac {1}{x}+\frac{1}{y}\geq \frac{4}{x+y} \) rezulta concluzia.
Vrajitoarea Andrei