Inegalitate nesimetrica

opincariumihai · Post by **opincariumihai** » Mon Jul 13, 2009 2:45 am

Daca \( a,b,c \) nenule si \( a^2+b^2+c^2=1 \) atunci :\( \frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}\leq \frac{3}{2}-(a-b)^2 \).

Mihai Opincariu

opincariumihai · Post by **opincariumihai** » Sun Sep 27, 2009 11:59 pm

Indicatie :
\( \frac{1}{2}- \frac{ab}{a^2+b^2} = \frac{(a-b)^2}{2(a^2+b^2)} \)

Virgil Nicula · Post by **Virgil Nicula** » Mon Sep 28, 2009 6:31 pm

opincariumihai wrote:\( \left\|\begin{array}{c}
\{a,b,c\}\subset\mathbb{R}^*\ \wedge\ a^2+b^2+c^2=1\\\\\\\\
m=\max\ \{\ |a-b|\ ,\ |b-c|\ ,\ |c-a|\ \}\end{array}\right\|\ \Longrightarrow\ \sum \frac{bc}{b^2+c^2}\leq \frac{3}{2}-m^2 \).

Demonstratie. Imi place problema ! De fapt inegalitatea propusa este simetrica.

Se arata usor ca \( \sum |b-c|\ge 2m\ (*) \) . Deci \( \sum \frac{bc}{b^2+c^2}=\frac 12\cdot\left(\sum\frac {2bc}{b^2+c^2}-1+1\right)= \)

\( \frac 12\cdot\left(3-\sum\frac {|b-c|^2}{b^2+c^2}\right)\ \stackrel {\mathrm{C.B.S.}}{\le}\ \frac 12\cdot\left[3-\frac {\left(\sum|b-c|\right)^2}{\sum\left(b^2+c^2\right)}\right]=\frac 32-\frac 14\cdot\left(\sum |b-c|\right)^2\ \stackrel{(*)}{ \le}\ \frac 32-m^2. \)