Inegalitate cu medii

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Marius Mainea
Gauss
Posts: 1077
Joined: Mon May 26, 2008 2:12 pm
Location: Gaesti (Dambovita)

Inegalitate cu medii

Post by Marius Mainea »

Daca \( x_1 ,x_2 ,x_3 ..... x_n \) sunt numere pozitive atunci:

\( x_1+x_2+.....+x_n+\frac{n}{\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n}}\ge (n+1)\sqrt[n]{x_1\cdot x_2....\cdot x_n} \)

MR , 6/2008
opincariumihai
Thales
Posts: 134
Joined: Sat May 09, 2009 7:45 pm
Location: BRAD

Post by opincariumihai »

Observ ca ati gasit o solutie foarte eleganta domnule Mainea ! Este vorba de solutia publicata aici , problema S 103.
Mai ramane insa de justificat (la fel de elegant) de ce are loc inegalitatea :
\( \frac{ \sum x_1...x_{n-1}}{n} \leq \( \frac{ x_1+...+x_n}{n}\)^{n-1} \)(*)
Last edited by opincariumihai on Sun Sep 27, 2009 10:41 pm, edited 1 time in total.
Marius Mainea
Gauss
Posts: 1077
Joined: Mon May 26, 2008 2:12 pm
Location: Gaesti (Dambovita)

Post by Marius Mainea »

Este inegalitatea Maclaurin:

Daca \( a_1,a_2...a_n \) sunt nenegative si\( S_k=\sqrt[k]{\frac{\sum a_1a_2...a_k}{C_n^k}} \) atunci \( S_1\ge S_2\ge...\ge S_n \).
opincariumihai
Thales
Posts: 134
Joined: Sat May 09, 2009 7:45 pm
Location: BRAD

Post by opincariumihai »

Alta mod pentru a justifica faptul ca \( \frac{ \sum x_1...x_{n-1}}{n} \leq \( \frac{ x_1+...+x_n}{n}\)^{n-1} \) ar fi sa folosim metoda lui Sturm.
Pentru suma constanta, membrul stang devine mai mare pentru \( x_1=x_2 \) etc.
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