Marius Dragoi wrote:Daca \( a_1,\ a_2,\ ...\ ,\ a_n \in R \), atunci sa se demonstreze urmatoarea inegalitate
\( \sum_{1 \leq i,j \leq n}^{} {\frac {a_i a_j}{i+j-1}}\ \geq\ 0 \).
O inegalitate dificila
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Marius Mainea
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O inegalitate dificila
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Marius Mainea
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Rezolvare de clasa a XII-a.
\( \sum _{i,j=1}^{n} {\frac{a_ia_j}{i+j-1}=\sum_{i,j=1}^n {a_ia_j\int_0^1 {t^{i+j-2}dt=\int_0^1 {(\sum_{i,j=1}^n {a_ia_jt^{i-1+j-1})dt= \)
\( =\int_0^1 {(\sum_{i=1}^n {a_{i}t^{i-1})^2}\geq (\int_0^1 {(\sum_{i=1}^n {a_it^{i-1}})dt})^2=(\sum {\frac{a_i}{i})^2\geq0 \)
unde am folosit inegalitatea CBS pentru integrale.
\( \sum _{i,j=1}^{n} {\frac{a_ia_j}{i+j-1}=\sum_{i,j=1}^n {a_ia_j\int_0^1 {t^{i+j-2}dt=\int_0^1 {(\sum_{i,j=1}^n {a_ia_jt^{i-1+j-1})dt= \)
\( =\int_0^1 {(\sum_{i=1}^n {a_{i}t^{i-1})^2}\geq (\int_0^1 {(\sum_{i=1}^n {a_it^{i-1}})dt})^2=(\sum {\frac{a_i}{i})^2\geq0 \)
unde am folosit inegalitatea CBS pentru integrale.
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Cezar Lupu
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Nu este nevoie de inegalitatea CBS pentru integrale pentru a rezolva inegalitatea asta. Pur si simplu, avem ca
\( \sum _{i,j=1}^{n} {\frac{a_ia_j}{i+j-1}=\sum_{i,j=1}^n {a_ia_j\int_0^1 {t^{i+j-2}dt=\int_0^1 {(\sum_{i,j=1}^n {a_ia_jt^{i-1+j-1})dt=\int_0^1 {(\sum_{i=1}^n {a_{i}t^{i-1})^2}\geq 0 \)
pentru ca integrala dintr-o functie pozitiva va fi, in mod evident, tot pozitiva.
\( \sum _{i,j=1}^{n} {\frac{a_ia_j}{i+j-1}=\sum_{i,j=1}^n {a_ia_j\int_0^1 {t^{i+j-2}dt=\int_0^1 {(\sum_{i,j=1}^n {a_ia_jt^{i-1+j-1})dt=\int_0^1 {(\sum_{i=1}^n {a_{i}t^{i-1})^2}\geq 0 \)
pentru ca integrala dintr-o functie pozitiva va fi, in mod evident, tot pozitiva.
Last edited by Cezar Lupu on Wed Jun 25, 2008 3:08 pm, edited 1 time in total.
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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Marius Mainea
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Cezar Lupu
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In legatura cu minorarea asta "mai buna" se poate da si alta solutie, dupa cum urmeaza:
Pentru orice \( x>0 \) definim functia
\( f(x)=\frac{1}{x}\left(\sum_{i=1}^{n}\frac{a_{i}x^i}{i}\right)^2-\sum_{i=1}^{n}\sum_{j=1}^{n}\frac{a_{i}a_{j}}{i+j-1} \).
Avem ca \( \lim_{x\to 0, x>0}f(x)=0 \) si \( f^{\prime}(x)=-\left(\sum_{i=1}^{n}(i-1)a_{i}x^{i-1}/i\right)^2\leq 0, \forall x>0 \). Rezulta astfel ca \( f \) este descrescatoare de unde obtinem ca \( f(1)\leq\lim_{x\to 0, x>0}f(x) \), deci concluzia problemei noastre. \( \qed \)
Observatie. Egalitatea are loc daca si numai daca \( a_{1}=a_{2}=\ldots =a_{n}=0 \).
Pentru orice \( x>0 \) definim functia
\( f(x)=\frac{1}{x}\left(\sum_{i=1}^{n}\frac{a_{i}x^i}{i}\right)^2-\sum_{i=1}^{n}\sum_{j=1}^{n}\frac{a_{i}a_{j}}{i+j-1} \).
Avem ca \( \lim_{x\to 0, x>0}f(x)=0 \) si \( f^{\prime}(x)=-\left(\sum_{i=1}^{n}(i-1)a_{i}x^{i-1}/i\right)^2\leq 0, \forall x>0 \). Rezulta astfel ca \( f \) este descrescatoare de unde obtinem ca \( f(1)\leq\lim_{x\to 0, x>0}f(x) \), deci concluzia problemei noastre. \( \qed \)
Observatie. Egalitatea are loc daca si numai daca \( a_{1}=a_{2}=\ldots =a_{n}=0 \).
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.