Fie \( a,b\in\mathbb{R} \) cu a<b si \( f:[a,b]\rightarrow\mathbb{R} \) o functie continua pe [a,b]. Aratati ca:
\( \int_a^bf(x)dx\cdot\int_a^b\sqrt{f^4(x)+1}dx\leq\frac{1}{2\sqrt{2}}(\int_a^b(f^2(x)+1)dx)^2 \)
Inegalitate integrala 2
Moderators: Bogdan Posa, Beniamin Bogosel, Marius Dragoi
-
Marius Mainea
- Gauss
- Posts: 1077
- Joined: Mon May 26, 2008 2:12 pm
- Location: Gaesti (Dambovita)
-
Cezar Lupu
- Site Admin
- Posts: 612
- Joined: Wed Sep 26, 2007 2:04 pm
- Location: Bucuresti sau Constanta
- Contact:
Solutie.
Se stie ca are loc urmatoarea descompunere \( x^4+1=(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1) \), de unde vom avea ca
\( \left(\int_a^b\sqrt{f^{4}(x)+1}dx\right)^{2}=\left(\int_a^b\sqrt{(f^{2}(x)-\sqrt{2}f(x)+1}\cdot\sqrt{f^{2}(x)+\sqrt{2}f(x)+1}dx\right)^{2} \).
Acum, folosind inegalitatea Cauchy-Schwarz, rezulta ca
\( \left(\int_a^b\sqrt{f^{4}(x)+1}dx\right)^{2}\leq\int_a^b (f^{2}(x)-\sqrt{2}f(x)+1)dx\cdot\int_a^b (f^{2}(x)+\sqrt{2}f(x)+1)dx=\left(\int_a^b (f^{2}(x)+1)dx\right)^{2}-\left(\int_a^b \sqrt{2}f(x)dx\right)^{2} \).
Acum, tot ce mai ramane de facut este sa aplicam inegalitatea elementara \( \alpha^2+\beta^2\geq 2\alpha\beta \) de unde vom avea ca
\( 2\int_a^b\sqrt{f^{4}(x)+1}dx\cdot\int_a^b\sqrt{2}f(x)dx\leq\left(\int_a^b\sqrt{f^{4}(x)+1}dx\right)^{2}+\left(\int_a^b\sqrt{2}f(x)\right)^{2} \). Combinand aceasta inegalitate cu cea de mai sus (dedusa din Cauchy-Schwarz), obtinem concluzia problemei. \( \qed \)
Se stie ca are loc urmatoarea descompunere \( x^4+1=(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1) \), de unde vom avea ca
\( \left(\int_a^b\sqrt{f^{4}(x)+1}dx\right)^{2}=\left(\int_a^b\sqrt{(f^{2}(x)-\sqrt{2}f(x)+1}\cdot\sqrt{f^{2}(x)+\sqrt{2}f(x)+1}dx\right)^{2} \).
Acum, folosind inegalitatea Cauchy-Schwarz, rezulta ca
\( \left(\int_a^b\sqrt{f^{4}(x)+1}dx\right)^{2}\leq\int_a^b (f^{2}(x)-\sqrt{2}f(x)+1)dx\cdot\int_a^b (f^{2}(x)+\sqrt{2}f(x)+1)dx=\left(\int_a^b (f^{2}(x)+1)dx\right)^{2}-\left(\int_a^b \sqrt{2}f(x)dx\right)^{2} \).
Acum, tot ce mai ramane de facut este sa aplicam inegalitatea elementara \( \alpha^2+\beta^2\geq 2\alpha\beta \) de unde vom avea ca
\( 2\int_a^b\sqrt{f^{4}(x)+1}dx\cdot\int_a^b\sqrt{2}f(x)dx\leq\left(\int_a^b\sqrt{f^{4}(x)+1}dx\right)^{2}+\left(\int_a^b\sqrt{2}f(x)\right)^{2} \). Combinand aceasta inegalitate cu cea de mai sus (dedusa din Cauchy-Schwarz), obtinem concluzia problemei. \( \qed \)
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.