Fie \( a,b,c \) numere reale pozitive. Sa se arate ca :
\( \frac{a^3}{b^2-bc+c^2}+\frac{b^3}{c^2-ca+a^2}+\frac{c^3}{a^2-ab+b^2}\ge \frac{3(ab+bc+ca)}{a+b+c}. \)
Inegalitate in trei variabile reale pozitive
Moderators: Laurian Filip, Beniamin Bogosel, Filip Chindea
Inegalitate in trei variabile reale pozitive
. A snake that slithers on the ground can only dream of flying through the air.
-
Marius Mainea
- Gauss
- Posts: 1077
- Joined: Mon May 26, 2008 2:12 pm
- Location: Gaesti (Dambovita)
Inegalitatea este echivalenta cu :
\( LHS+2(a+b+c)\ge RHS+2(a+b+c) \) sau
\( (a^3+b^3+c^3)\sum {\frac{1}{a^2-ab+b^2}}\ge\frac{2(a^2+b^2+c^2)+7(ab+bc+ca)}{a+b+c} \)
Insa folsind AM-HM
\( (a^3+b^3+c^3)\sum {\frac{1}{a^2-ab+b^2}}\ge \frac{9(a^3+b^3+c^3)}{2(a^2+b^2+c^2)-(ab+bc+ca)}\ge \frac{2(a^2+b^2+c^2)+7(ab+bc+ca)}{a+b+c} \) ultima inegalitate reducandu-se la
\( 5\sum_{cyc} a^4+2\sum_{cyc}a^2bc\ge\sum_{cyc}a^2b^2+3\sum_{cyc}a^3(b+c) \) care este adevarata.
\( LHS+2(a+b+c)\ge RHS+2(a+b+c) \) sau
\( (a^3+b^3+c^3)\sum {\frac{1}{a^2-ab+b^2}}\ge\frac{2(a^2+b^2+c^2)+7(ab+bc+ca)}{a+b+c} \)
Insa folsind AM-HM
\( (a^3+b^3+c^3)\sum {\frac{1}{a^2-ab+b^2}}\ge \frac{9(a^3+b^3+c^3)}{2(a^2+b^2+c^2)-(ab+bc+ca)}\ge \frac{2(a^2+b^2+c^2)+7(ab+bc+ca)}{a+b+c} \) ultima inegalitate reducandu-se la
\( 5\sum_{cyc} a^4+2\sum_{cyc}a^2bc\ge\sum_{cyc}a^2b^2+3\sum_{cyc}a^3(b+c) \) care este adevarata.