Aratati ca pentru orice \( a,b,c>0 \) are loc \( \sum_{cyc}\frac{a^2+bc}{b+c}\ge a+b+c \).
Bogdan Enescu, OJ Bacau, 2000
OJ Bacau 2000
Moderators: Laurian Filip, Beniamin Bogosel, Filip Chindea
OJ Bacau 2000
. A snake that slithers on the ground can only dream of flying through the air.
\( \sum \frac{a^2+bc}{b+c}\geq\sum a\Leftrightarrow \sum \frac{(a-b)(a-c)}{b+c}\geq0\Leftrightarrow\sum (a^2-b^2)(a^2-c^2)\geq0 \)
si putem presupune \( a\geq b\geq c \) caci e simetrica si apoi se transforma echivalent in:
\( (a^2-c^2)^2\geq(b^2-c^2)(a^2-b^2) \) care e evidenta. Egalitate cand \( a=b=c \).
si putem presupune \( a\geq b\geq c \) caci e simetrica si apoi se transforma echivalent in:
\( (a^2-c^2)^2\geq(b^2-c^2)(a^2-b^2) \) care e evidenta. Egalitate cand \( a=b=c \).
n-ar fi rau sa fie bine 