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O.N.M. 1997; M.B.
Posted: Fri Jan 02, 2009 1:08 pm
by Claudiu Mindrila
Problema. Fie \( x,y,z \) numere pozitive astfel incat \( xyz=1 \). Sa se arate ca \( \sum \frac{x^9+y^9}{x^6+x^3y^3+y^6} \geq 2 \).
Mircea Becheanu, O.N.M. 1997
Posted: Fri Jan 16, 2009 11:42 pm
by Marius Mainea
Notand \( a=x^3 \) , \( b=y^3 \) \( c=z^3 \) inegalitatea se reduce la
\( \sum {\frac{a^3+b^3}{a^2+ab+b^2}}\ge 2 \) cu \( abc=1 \)
Insa \( LHS\ge\sum \frac{(a^2+b^2)(a+b)}{2(a^2+\frac{a^2+b^2}{2}+b^2)}=\sum{\frac{(a^2+b^2)(a+b)}{3(a^2+b^2)}}=\frac{2}{3}\sum a\ge2\sqrt[3]{abc}=RHS \)