Fie \( f:[1,2]\rightarrow (0,\infty) \) o functie strict crescatoare, derivabila cu derivata continua si \( f(2)=2f(1) \). Demonstrati inegalitatea urmatoare si precizati cazul de egalitate:
\( \int_1^2\frac{f^\prime(x)}{xf^\prime(x)+f(x)}dx\leq\frac{1}{2}\ln2 \).
Inegalitate integrala 1
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Marius Mainea
- Gauss
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Marius Mainea
- Gauss
- Posts: 1077
- Joined: Mon May 26, 2008 2:12 pm
- Location: Gaesti (Dambovita)
\( f \) crescatoare \( \Rightarrow f^\prime \geq0 \)
Din inegalitatea mediilor avem: \( \frac{f^\prime(x)}{xf^\prime(x)+f(x)}= \frac{2}{x+\frac{f(x)}{f^\prime(x)}} \leq\frac{1}{2}(\frac{1}{x}+\frac{f^\prime(x)}{f(x)}) \)
Integrand inegalitatea obtinem :
\( \int_1^2\frac{f^\prime(x)}{xf^\prime(x)+f(x)}dx\leq\frac{1}{2}\int_1^2\frac{1}{2}(\frac{1}{x}+\frac{f^\prime(x)}{f(x)})dx
\Leftrightarrow\int_1^2\frac{f^\prime(x)}{xf^\prime(x)+f(x)}dx\leq\frac{1}{4}[\ln(2)+\ln(2f(1))-\ln(f(1)] \)
\( \Leftrightarrow\int_1^2\frac{f^\prime(x)}{xf^\prime(x)+f(x)}dx\leq\frac{1}{2}\ln(2) \)
Din inegalitatea mediilor avem: \( \frac{f^\prime(x)}{xf^\prime(x)+f(x)}= \frac{2}{x+\frac{f(x)}{f^\prime(x)}} \leq\frac{1}{2}(\frac{1}{x}+\frac{f^\prime(x)}{f(x)}) \)
Integrand inegalitatea obtinem :
\( \int_1^2\frac{f^\prime(x)}{xf^\prime(x)+f(x)}dx\leq\frac{1}{2}\int_1^2\frac{1}{2}(\frac{1}{x}+\frac{f^\prime(x)}{f(x)})dx
\Leftrightarrow\int_1^2\frac{f^\prime(x)}{xf^\prime(x)+f(x)}dx\leq\frac{1}{4}[\ln(2)+\ln(2f(1))-\ln(f(1)] \)
\( \Leftrightarrow\int_1^2\frac{f^\prime(x)}{xf^\prime(x)+f(x)}dx\leq\frac{1}{2}\ln(2) \)