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
Inegalitate polinomiala 2
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Marius Mainea
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opincariumihai
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\( 3(a^2b^2+b^2c^2+c^2a^2)(a^2+b^2+c^2)=3a^2b^2c^2+3\prod(a^2+b^2) \)
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 !
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 !