Disuguaglianza iraniana

Polinomi, disuguaglianze, numeri complessi, ...
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jordan
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Disuguaglianza iraniana

Messaggio da jordan » 20 feb 2013, 23:39

Siano $a,b,c$ reali positivi; mostrare che \[ \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{(a+b+c)^2}\ge\frac{7}{25}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a+b+c}\right)^2 \]
The only goal of science is the honor of the human spirit.

totissimus
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Re: Disuguaglianza iraniana

Messaggio da totissimus » 23 feb 2013, 12:03

$ \displaystyle a\geq b\geq c $

$ \displaystyle \frac{28}{9a^2}=\frac{3}{a^2}+\frac{1}{(3a)^2}=\frac{7}{25}\left( \frac{3}{a}+\frac{1}{3a}\right)^2\leq \frac{7}{25}\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a+b+c} \right)^2$.

La funzione $\displaystyle \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{(a+b+c)^2}$ rispetto a $c$ è decrescente in $[b,\infty)$ quindi:

$ \displaystyle \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{(a+b+c)^2}\leq \frac{1}{a^2}+\frac{2}{b^2}+\frac{1}{(a+2b)^2}$
La funzione di $b$: $ \displaystyle \frac{1}{a^2}+\frac{2}{b^2}+\frac{1}{(a+2b)^2}$ è decrescente in $[a,\infty)$

quindi:

$\displaystyle \frac{1}{a^2}+\frac{2}{b^2}+\frac{1}{(a+2b)^2}\leq \frac{3}{a^2}+\frac{1}{(3a)^2}= \frac{28}{9a^2}$

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Re: Disuguaglianza iraniana

Messaggio da jordan » 25 feb 2013, 01:14

totissimus ha scritto:$ \displaystyle a\geq b\geq c $

$ \displaystyle \frac{28}{9a^2}=\frac{3}{a^2}+\frac{1}{(3a)^2}=\frac{7}{25}\left( \frac{3}{a}+\frac{1}{3a}\right)^2\leq \frac{7}{25}\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a+b+c} \right)^2$.
Fin qui ok.
totissimus ha scritto:La funzione $\displaystyle \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{(a+b+c)^2}$ rispetto a $c$ è decrescente in $[b,\infty)$ quindi:

$ \displaystyle \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{(a+b+c)^2}\leq \frac{1}{a^2}+\frac{2}{b^2}+\frac{1}{(a+2b)^2}$
Se imponi wlog $a \ge b \ge c > 0 $ allora $ c \in (0,b]$ e quella funzione non ha limite superiore..
The only goal of science is the honor of the human spirit.

totissimus
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Re: Disuguaglianza iraniana

Messaggio da totissimus » 02 mar 2013, 16:24

Ho proprio scritto una emerita castroneria!!

Rimedio con un altro tentativo.

$\displaystyle \triangle=\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{(a+b+c)^{2}}-\frac{7}{25}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a+b+c}\right)^{2}$

$\displaystyle =\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{(a+b+c)^{2}}-\frac{7}{25}\left[\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+2\frac{1}{a+b+c}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\right]$

$\displaystyle =\frac{18}{25}\left[\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{(a+b+c)^{2}}\right]-\frac{14}{25}\frac{1}{a+b+c}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$

$\displaystyle \frac{25\triangle}{2}=9\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right)+\frac{9}{(a+b+c)^{2}}-\frac{7}{a+b+c}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$

$\displaystyle =\left[\frac{3}{a+b+c}-\frac{7}{6}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\right]^{2}+9\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right)-\frac{49}{36}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2}$

Sussiste la disuguaglianza: $\displaystyle (x+y+z)^2\leq 3(x^2+y^2+z^2)$ e quindi:

$\displaystyle \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2}\leq3\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right)$

da cui:
$\displaystyle \frac{25\triangle}{2}\geq\left[\frac{3}{a+b+c}-\frac{7}{6}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\right]^{2}+9\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right)-\frac{49}{12}\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right)$

$\displaystyle =\left[\frac{3}{a+b+c}-\frac{7}{6}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\right]^{2}+\frac{59}{12}\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right)\geq0$

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