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Inegalitati de tip Aczel :


Daca \( a \) , \( b \) , \( c \) sunt trei numere reale positive, atunci :

\( \odot\ \ \underline {\overline {\left\|\ (ab + bc + ca)\sqrt 3 + \frac 12\cdot \sum (b - c)^2\ \le\ (a + b + c)\cdot\sqrt {a^2 + b^2 + c^2}\ \right\|}}\ . \)

\( \odot\ \ 3\sqrt {abc(a+b+c)}+\frac {\sqrt 2}{2}\cdot \sum (b-c)^2\le \left(a^2+b^2+c^2\right)\sqrt 3\ . \)

\( \odot\ \ 3(ab+bc+ca)+\frac {\sqrt 2}{2}\cdot \sum (b-c)^2\le (a+b+c)\cdot\sqrt {3\left(a^2+b^2+c^2\right)}\ . \)


Daca \( 0 < a_1\le a_2\le \ldots \le a_n \) si \( x_1\ge x_2\ge \ldots\ge x_n>0 \) atunci

\( \underline {\overline {\left\|\ \frac 1n\cdot\sqrt {\sum_{1\le j<k\le n} (a_j-a_k)^2\cdot\sum_{1\le j<k\le n} (x_j-x_k)^2} + \sum_{k=1}^n a_kx_k\ \le\ \sqrt {\sum_{k=1}^n a_k^2\cdot\sum_{k=1}^n x_k^2}\ \right\|}} \)

care este o intarire a inegalitatii C.B.S. \( \sum_{k=1}^n a_kx_k\ \le\ \sqrt {\sum_{k=1}^n a_k^2\cdot\sum_{k=1}^n x_k^2} \)

in situatia speciala cand cele doua siruri finite \( a_k \) , \( x_k \) , \( k\in\overline {1,n} \) sunt invers ordonate.

Virgil Nicula wrote:Inegalitati de tip Aczel :


Daca \( a \) , \( b \) , \( c \) sunt trei numere reale positive, atunci :

\( \odot\ \ \underline {\overline {\left\|\ (ab + bc + ca)\sqrt 3 + \frac 12\cdot \sum (b - c)^2\ \le\ (a + b + c)\cdot\sqrt {a^2 + b^2 + c^2}\ \right\|}}\ . \)

Notam x=a+b+c si y=ab+bc+ca si inegalitatea se reduce la

\( y\sqrt{3}+x^2-3y\le x\sqrt{x^2-2y} \) si prin ridicare la patrat si reducerea termenilor asemenea , simplificari etc,

\( 3(\sqrt{3}-1)^2(2+\sqrt{3})y\le 2x^2 \)

sau \( 3y\le x^2 \)

ceea ce este evident.

Virgil Nicula wrote:
\( \odot\ \ 3\sqrt {abc(a+b+c)}+\frac {\sqrt 2}{2}\cdot \sum (b-c)^2\le \left(a^2+b^2+c^2\right)\sqrt 3\ . \)

Daca \( x=a+b+c , y=ab+bc+ca , z=abc \)

inegalitatea devine \( 3\sqrt{xz}+ \sqrt{2}(x^2-3y)\le \sqrt{3}(x^2-2y) \)

Dar cum \( \sqrt{xz}\le \frac{y}{\sqrt{3}} \)

\( LHS\le \sqrt{2}x^2+(\sqrt{3}-3\sqrt{2})y\le \sqrt{3}(x^2-2y) \Longleftrightarrow 3y\le x^2 \)

Virgil Nicula wrote:
\( \odot\ \ 3(ab+bc+ca)+\frac {\sqrt 2}{2}\cdot \sum (b-c)^2\le (a+b+c)\cdot\sqrt {3\left(a^2+b^2+c^2\right)}\ . \)

analog .

Frumos, profesor Mainea ! Ma bucur ca problemele mele v-au trezit interesul sa le abordati.
Si ultima, se incumeta cineva ?! Toate patru inclusiv Aczel au acelasi izvor, foarte simplu. Mai astept.

Virgil Nicula wrote:Inegalitati de tip Aczel :

Daca \( 0 < a_1\le a_2\le \ldots \le a_n \) si \( x_1\ge x_2\ge \ldots\ge x_n>0 \) atunci

\( \underline {\overline {\left\|\ \frac 1n\cdot\sqrt {\sum_{1\le j<k\le n} (a_j-a_k)^2\cdot\sum_{1\le j<k\le n} (x_j-x_k)^2} + \sum_{k=1}^n a_kx_k\ \le\ \sqrt {\sum_{k=1}^n a_k^2\cdot\sum_{k=1}^n x_k^2}\ \right\|}} \)

care este o intarire a inegalitatii C.B.S. \( \sum_{k=1}^n a_kx_k\ \le\ \sqrt {\sum_{k=1}^n a_k^2\cdot\sum_{k=1}^n x_k^2} \)

in situatia speciala cand cele doua siruri finite \( a_k \) , \( x_k \) , \( k\in\overline {1,n} \) sunt invers ordonate.

Notam: \( A=\displaystyle\sum^n_{k=1}a_{k}^2, \) \( X=\displaystyle\sum^n_{k=1}x_{k}^2, \) \( ax=\displaystyle\sum^n_{k=1}a_{k}x_{k}, \) \( S_{a}=\displaystyle\sum^n_{k=1}a_{k}, \) si \( S_{x}=\displaystyle\sum^n_{k=1}x_{k}. \)

Inegalitatea este echivalenta cu: \( \sqrt{AX}\geq ax+\frac{1}{n}\sqrt{(nA-S_{a}^2)(nX-S_{x}^2)}. \)

Din Inegalitatea lui Aczel obtinem: \( (nA-S_{a}^2)(nX-S_{x}^2)\leq (n\cdot\sqrt{AX}-S_{a}\cdot S_{x})^2, \) de unde

\( \frac{1}{n}\sqrt{(nA-S_{a}^2)(nX-S_{x}^2)}\leq\frac{1}{n}\cdot (n\sqrt{AX}-S_{a}S_{x})=\sqrt{AX}-\frac{S_{a}S_{x}}{n}. \)

Mai ramane de aratat ca: \( S_{a}S_{x}\geq n\cdot ax, \) adica: \( (a_{1}+a_{2}+\dots+a_{n})(x_{1}+x_{2}+\dots+x_{n})\geq n(a_{1}x_{1}+a_{2}x_{2}+\dots+a_{n}x_{n}), \) care este chiar

inegalitatea lui Cebasev cand \( n \)-trupletele \( (a_{1},a_{2}\dots, a_{n}) \) si \( (x_{1},x_{2}\dots, x_{n}) \) au monotonii diferite.