Fie \( m,\ n\in\mathbb{N}^{*},\ m\le2n \) si \( a,\ b,\ c>0. \)Sa se arate ca: \( \frac{a^{m}}{b^{n}+c^{n}}+\frac{b^{m}}{c^{n}+a^{n}}+\frac{c^{m}}{a^{n}+b^{n}}\ge\frac{3}{2}\sqrt{\frac{a^{m}+b^{m}+c^{m}}{a^{2n-m}+b^{2n-m}+c^{2n-m}}}. \)
Claudiu Mindrila, Gazeta Matematica nr. 6/2010
Inegalitate din G. M. 6/2010
Moderators: Filip Chindea, Andrei Velicu, Radu Titiu
-
Claudiu Mindrila
- Fermat
- Posts: 520
- Joined: Mon Oct 01, 2007 2:25 pm
- Location: Targoviste
- Contact:
Inegalitate din G. M. 6/2010
elev, clasa a X-a, C. N. "C-tin Carabella", Targoviste
-
Mateescu Constantin
- Newton
- Posts: 307
- Joined: Tue Apr 21, 2009 8:17 am
- Location: Pitesti
\( \odot\ \ LHS\ \stackrel{\small\mbox{CEB}}{\ge}\ \frac 13\cdot\left(a^m+b^m+c^m\right)\left(\frac 1{a^n+b^n}+\frac 1{b^n+c^n}+\frac 1{c^n+a^n}\right)\ \stackrel{\small\mbox{C.B.S.}}{\ge}\ \frac 32\ \cdot\ \frac {a^m+b^m+c^m}{a^n+b^n+c^n} \) .
\( \odot\ \ \) Ramane sa aratam inegalitatea \( \frac{a^m+b^m+c^m}{a^n+b^n+c^n}\ \ge\ \sqrt{\frac {a^m+b^m+c^m}{a^{2n-m}+b^{2n-m}+c^{2n-m}}} \) care se scrie :
\( (a^m+b^m+c^m)(a^{2n-m}+b^{2n-m}+c^{2n-m})\ \stackrel{\small\mbox{C.B.S.}}{\ge}\ (a^n+b^n+c^n)^2=\left\(\sum\ a^{\frac {2n-m}2}\ \cdot\ a^{\frac m2}\right\)^2 \)
\( \odot\ \ \) Ramane sa aratam inegalitatea \( \frac{a^m+b^m+c^m}{a^n+b^n+c^n}\ \ge\ \sqrt{\frac {a^m+b^m+c^m}{a^{2n-m}+b^{2n-m}+c^{2n-m}}} \) care se scrie :
\( (a^m+b^m+c^m)(a^{2n-m}+b^{2n-m}+c^{2n-m})\ \stackrel{\small\mbox{C.B.S.}}{\ge}\ (a^n+b^n+c^n)^2=\left\(\sum\ a^{\frac {2n-m}2}\ \cdot\ a^{\frac m2}\right\)^2 \)