Daca functia \( f:[0,1]\to\mathbb{R} \) este continua, sa se arate ca
\( \int_0^1f^{2}(x^{2})dx\geq\frac{3}{4}\left(\int_0^1f(x)dx\right)^{2} \).
Laurentiu Panaitopol, Olimpiada Judeteana Giurgiu, 1991
Inegalitate integrala cu o functie continua
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Cezar Lupu
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Inegalitate integrala cu o functie continua
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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Edgar Dobriban
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Din inegalitatea Cauchy avem \( \int_0^1f^{2}(x^{2})dx\int_0^1(2x)^2dx\geq\left(\int_0^1f(x^2)2xdx\right)^{2} \).Cezar Lupu wrote:Daca functia \( f:[0,1]\to\mathbb{R} \) este continua, sa se arate ca
\( \int_0^1f^{2}(x^{2})dx\geq\frac{3}{4}\left(\int_0^1f(x)dx\right)^{2} \).
Laurentiu Panaitopol, Olimpiada Judeteana Giurgiu, 1991
Dar \( \int_0^1f(x^2)2xdx=\int_0^1f(x)dx \) si \( \int_0^1(2x)^2dx=4/3 \) de unde rezulta concluzia.