Inecuatie in Z
Posted: Thu Dec 04, 2008 8:11 pm
Rezolvati in \( \mathbb{Z} \) inecuatia \( a^2+b^2+2c^2+2bc+a+b\le 2ac. \)
Gheorghe Ghita, RMT 4/2008
Gheorghe Ghita, RMT 4/2008
INDICATIE:
\(
\mbox{Avem: }a^2+b^2+2c^2+2bc+a+b\le 2ac\Leftrightarrow (a-c)^2+(b+c)^2\le -a-b.\ (1)\\
\mbox{Sa mai observam apoi ca: } n^2\ge n; (\forall)n\in\mathbb{Z}. \mbox{ Asa ca: } \\
\left \begin{\array} (a-c)^2\ge c-a; \ (\forall)a;b\in\mathbb{Z} \\
(b+c)^2\ge -(b+c); \ (\forall)b;c\in\mathbb{Z}\right\}\Rightarrow (a-c)^2+(b+c)^2\ge -a-b. \ (2)
\)