Fie \( f:[0,1]\to [0, \infty) \) o functie integrabila. Sa se arate ca
\( \int_0^1 f(x)dx\cdot\int_0^1f^{2}(x)dx\leq\int_0^1f^{3}(x)dx \).
Laurentiu Panaitopol, locala Bucuresti 2005
Inegalitate integrala gen Muirhead
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Cezar Lupu
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Inegalitate integrala gen Muirhead
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.
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Doru Popovici
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Folosim CBS de doua ori:
\( \int_{0}^{1}f^3(x)dx\int_{0}^{1}f(x)dx=\int_{0}^{1}(f(x)sqrt{f(x)})^2dx \int_{0}^{1} (sqrt{f(x)})^2dx\geq(\int_{0}^{1}f^2(x)dx)^2= \)
\( \int_{0}^{1}f^2(x)dx\int_{0}^{1}f^2(x)dx\int_{0}^{1}dx\geq \int_{0}^{1}f^2(x)dx(\int_{0}^{1} f(x)dx)^2. \)
Daca \( \int_{0}^{1} f(x)dx>0 \), atunci rezulta inegalitatea.
Daca \( \int_{0}^{1} f(x)dx=0 \), atunci rezulta \( \int_{0}^{1} f^3(x)dx\geq 0 \).
\( \int_{0}^{1}f^3(x)dx\int_{0}^{1}f(x)dx=\int_{0}^{1}(f(x)sqrt{f(x)})^2dx \int_{0}^{1} (sqrt{f(x)})^2dx\geq(\int_{0}^{1}f^2(x)dx)^2= \)
\( \int_{0}^{1}f^2(x)dx\int_{0}^{1}f^2(x)dx\int_{0}^{1}dx\geq \int_{0}^{1}f^2(x)dx(\int_{0}^{1} f(x)dx)^2. \)
Daca \( \int_{0}^{1} f(x)dx>0 \), atunci rezulta inegalitatea.
Daca \( \int_{0}^{1} f(x)dx=0 \), atunci rezulta \( \int_{0}^{1} f^3(x)dx\geq 0 \).