Inegalitate ...numere reale

lflorea Joi Dec 13, 2012 1:06 am

Aratati ca daca a, b, c sunt numere reale cu proprietatea ca
$abc(a^{2}+b^{2}+c^{2})+\left ( 3-abc \right )\left ( ab+bc+ca \right )=0$
atunci $abc\leqslant 1$

**Admin** Joi Dec 13, 2012 3:06 am

$\begin{array}{*{20}{l}} {abc\sum {{a^2} + \left( {3 - abc} \right)\sum {ab} } = 0}\\ {}\\ {{\text{Pentru }}a = b = c = 0 \Rightarrow abc = 0 < 1 \to {\text{Inegalitate adevarata!}}}\\ {}\\ {{\text{Presupunem ca }}abc \ne 0,{\mkern 1mu} {\rm{atunci:}}}\\ {}\\ {abc\sum {{a^2} = \left( {abc - 3} \right)\sum {ab} } \Leftrightarrow \sum {{a^2} = \frac{{\left( {abc - 3} \right)}}{{abc}}\sum {ab} } }\\ {}\\ {{{\left( {MA - MG} \right)}_2}:}\\ {}\\ {\sum {{a^2}} \ge \sum {ab} \Rightarrow \frac{{\left( {abc - 3} \right)}}{{abc}}\sum {ab} \ge \sum {ab} \Leftrightarrow }\\ {}\\ {\left( {\sum {ab} } \right)\left[ {\frac{{\left( {abc - 3} \right) - abc}}{{abc}}} \right] \ge 0 \Leftrightarrow - \left( {\sum {ab} } \right)\frac{3}{{abc}} \ge 0\ \Rightarrow \left\{ {\begin{array}{*{20}{l}} {abc < 0 < 1}\\ {\sum {ab > 0} } \end{array}} \right. \vee \left\{ {\begin{array}{*{20}{l}} {abc > 0}\\ {\sum {ab < 0} } \end{array}} \right.} \end{array}$

De aici rămâne de discutat cele 2 cazuri .... primul caz se demonstrează extrem de simplu !

Inegalitate ...numere reale

Inegalitate ...numere reale

Re: Inegalitate ...numere reale