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Fie \( a,b,c \) numere reale strict pozitive cu \( abc=1 \). Sa se arate ca \( \frac{a^{10}}{b+c}+\frac{b^{10}}{c+a}+\frac{c^{10}}{a+b} \geq \frac{a^7}{b^7+c^7}+\frac{b^7}{c^7+a^7}+\frac{c^7}{a^7+b^7} \).
Claudiu Mindrila

\( \sum {\frac{a^{10}}{b+c}}=\sum {\frac{a^7}{b^3c^3(b+c)}}\ge\sum {\frac{a^7}{(\frac{b+c}{2})^6(b+c)}}=\sum {\frac{a^7}{\frac{(b+c)^7}{2^6}}\ge\sum {\frac{a^7}{b^7+c^7}} \)

conform inegalitatii mediilor (forma slaba si forma tare)

Marius Mainea wrote:\( \sum {\frac{a^7}{\frac{(b+c)^7}{2^6}}\geq\sum {\frac{a^7}{b^7+c^7}} \)

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Claudiu Mindrila wrote:

Marius Mainea wrote:\( \sum {\frac{a^7}{\frac{(b+c)^7}{2^6}}\geq\sum {\frac{a^7}{b^7+c^7}} \)

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\( \left(\frac{a_1+a_2+.....+a_n}{n}\right)^k\le\frac{a_1^k+a_2^k+...+a_n^k}{n} \) , \( a_i \) pozitive.(Inegalitatea,,tare'' a mediilor)

Claudiu, cu putina rabdare, urmand aceeasi cale de demonstratie o puteai extinde astfel :

Claudiu Mindrila wrote:Fie \( \{a,b,c\}\subset \mathbb R^*_{+} \) pentru care \( abc=1 \) si \( \{m,n,p}\subset \mathbb N^* \) . Sa se arate ca

\( \frac{a^{m+n}}{(b+c)^{p}}+\frac{b^{m+n}}{(c+a)^{p}}+\frac{c^{m+n}}{(a+b)^{p}} \geq \frac {1}{2^{p-1}}\cdot\left(\frac{a^m}{b^{2n+p}+c^{2n+p}}+\frac{b^m}{c^{2n+p}+a^{2n+p}}+\frac{c^m}{a^{2n+p}+b^{2n+p}}\right) \) .

Claudiu Mindrila

Consider ca si aceasta extindere iti apartine. Eu nu am facut mare lucru. Ideea de fapt iti apartine.

Caz particular : \( m=7 \) , \( n=3 \) , \( p=1 \) . Draguta inegalitate ! Merge bine si la clasa a VII - a

daca acestia stiu ca \( a>0 \) , \( b>0 \) , \( n\in \mathbb N^*\ \Longrightarrow\ (a+b)^n\le 2^{n-1}\cdot\left(a^n+b^n\right) \) \( \ (*) \) .

Iata aici o demonstratie. Deoarece \( (a-b)(X-Y)\ge 0\ \Longleftrightarrow\ (a+b)(X+Y)\le 2\cdot (aX+bY) \)

atunci urmatoarele inegalitati devin evidente si produsul lor este \( (*) \) .

\( (a+b)^2\le 2\cdot\left(a^2+b^2\right) \)

\( (a+b)\left(a^2+b^2\right)\le 2\cdot\left(a^3+b^3\right) \)

\( (a+b)\left(a^3+b^3\right)\le 2\cdot\left(a^4+b^4\right) \)
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\( (a+b)\left(a^{n-2}+b^{n-2}\right)\le 2\cdot\left(a^{n-1}+b^{n-1}\right) \)

\( (a+b)\left(a^{n-1}+b^{n-1}\right)\le 2\cdot\left(a^{n}+b^{n}\right) \)

Procedand asemanator poti demonstra asa zisa "inegalitate tare a mediilor" cum zice prof. Marius Mainea.

Eu stiam ca ea este un caz particular (linii identice) al inegalitatii Cebasev extinsa la o matrice \( A=\left(a_{ij}\right)\in \mathbb M_{mn} \)

cu toate liniile crescatoare de termeni pozitivi : \( \prod_{i=1}^m\sum_{j=1}^na_{ij}\le n^{m-1}\cdot\sum_{j=1}^n\prod_{i=1}^ma_{ij} \) .

Claudiu, cam exagerezi cu atatea inrgalitati. Posteaza-le si tu la o distanta mai mare in timp.