Demonstrarea unei inegalitati. Rezolvata

Nikolle Dum Aug 26, 2012 4:58 pm

As vrea sa ma ajutati la aceste 2 puncte:

a) Aratati ca $a^{4}+b^{4}+c^{4}\geq a^{2}\ast b^{2} +b^{2}\ast c^{2}+c^{2}\ast a^{2}, \bigtriangledown a,b,c\epsilon \mathbb{R}$
b) $a^{4}+b^{4}+c^{^{4}} \geq abc(a+b+c),\bigtriangledown a,b,c \epsilon \mathbb{R}$

**Admin** Dum Aug 26, 2012 6:37 pm

a) Aratati ca $a^{4}+b^{4}+c^{4}\geq a^{2}\ast b^{2} +b^{2}\ast c^{2}+c^{2}\ast a^{2}, \bigtriangledown a,b,c\epsilon \mathbb{R}$

$\begin{array}{l} {a^4} + {b^4} + {c^4} \ge {a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2}\\ {\text{Din }}{\left( {{\text{MA}} - {\text{MG}}} \right)_2}:\\ \left\{ \begin{array}{l} {a^4} + {b^4} \ge 2\sqrt {{a^4}{b^4}} = 2{a^2}{b^2}\\ {a^4} + {c^4} \ge 2\sqrt {{a^4}{c^4}} = 2{a^2}{c^2}\\ {b^4} + {c^4} \ge 2\sqrt {{b^4}{c^4}} = 2{b^2}{c^2} \end{array} \right. \Rightarrow 2\left( {{a^4} + {b^4} + {c^4}} \right) \ge 2\left( {{a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2}} \right) \Rightarrow \\ {a^4} + {b^4} + {c^4} \ge {a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2}\\ {\text{Sau:}}\\ \sum {{a^4}} \ge \sum {{a^2}{b^2}} \to {\text{alta notatie matematica}} \end{array}$
Problema b tema !

Demonstrarea unei inegalitati. Rezolvata

Demonstrarea unei inegalitati. Rezolvata

Re: Demonstrarea unei inegalitati. Rezolvata