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Inegalitate polinomiala 2
Posted: Sat Jun 06, 2009 2:42 pm
by Marius Mainea
Daca a,b,c sunt numere pozitive atunci :
\( 3(a^2b^2+b^2c^2+c^2a^2)(a^2+b^2+c^2)\ge (a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2) \)
M.R. 3/2009
Posted: Sat Aug 08, 2009 11:18 am
by opincariumihai
\( 3(a^2b^2+b^2c^2+c^2a^2)(a^2+b^2+c^2)=3a^2b^2c^2+3\prod(a^2+b^2) \)
Impartind cu
\( a^2b^2c^2 \) si notand
\( x=a/b... \) inegalitatea de demonstrat se scrie echivalent
\( 3+3\prod{(x^2+1)}\geq\prod{(x^2+x+1)} \) unde
\( xyz=1 \) .Notand cu
\( A=x+y+z \) si
\( B=xy+yz+zx \) , dupa efectuarea calculelor , inegalitatea de demonstrat devine :
\( 9+2A^2+2B^2\geq6A+6B+AB \) adica
\( (A-B)^2+(A-3)^2+(B-3)^2+AB-9\geq0 \) inegalitate evidenta daca tinem cont ca
\( AB-9=(x+y+z)(xy+yz+zx)-9xyz\geq0 \) 
P.S. Sunt curios cum ar arata o solutie fara calcule !