Question

Va roog(cred ca e ceva legat de inegalitatea mediilor...)

Answer 1

$\displaystyle Este~vorba~despre~binecunoscuta~inegalitate~ \\ \\ a^2+b^2+c^2 \geq \frac{(a+b+c)^2}{3}. \\ \\ ------------------------------ \\ \\ Poate~fi~demonstrata~in~mai~multe~moduri: \\ \\ 1.~Inegalitatea~este~echivalenta~cu~ \\ \\ 3(a^2+b^2+c^2) \geq a^2+b^2+c^2 +2ab+2bc+2ac \Leftrightarrow \\ \\ \Leftrightarrow 2(a^2+b^2+c^2) \geq 2ab+2bc+2ac \Leftrightarrow \\ \\ \Leftrightarrow (a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ac+c^2) \geq 0 \Leftrightarrow$
$\displaystyle \Leftrightarrow (a-b)^2+(b-c)^2+(c-a)^2 \geq 0,~evident~adevarata,~caci \\ \\ x^2 \geq 0~\forall~x \in \mathbb{R}. \\ \\ 2.~ Inegalitatea~a^2+b^2+c^2 \geq ab+bc+ac ~mai~poate~fi~\\ \\demonstrata~ astel: \\ \\ Din~inegalitatea~mediilor~avem~x^2+y^2 \geq 2|x||y| \geq2xy. \\ \\ Rezulta~totodata~din~faptul~ca~(x-y)^2 \geq 0.~ \\ \\ a^2+b^2 \geq 2ab \\ \\ b^2+c^2 \geq 2bc \\ \\ a^2+c^2 \geq 2ac \\ \\ Prin~insumare~obtinem~2(a^2+b^2+c^2) \geq 2(ab+bc+ac).$
$\displaystyle 3.~O~alta~metoda:~Aplicam~inegalitatea~Cauchy-Buniakovski- \\ \\ Schwartz. \\ \\ \frac{a^2}{1}+ \frac{b^2}{1}+ \frac{c^2}{1} \geq \frac{(a+b+c)^2}{1+1+1}. \\ \\ 4.~Se~poate~totodata~observa~ca~inegalitatea~provine~din~inegalitatea \\ \\ rearanjarii.$