Concursul ,,Marian Tarina'', 2008

[Marius Mainea]

Sa se arate ca pentru orice \( x,y,z \in \mathbb{R^{\ast} \) are loc inegalitatea:

\( \frac{x^2}{x^{12}+y^6+z^6}+\frac{y^2}{x^6+y^{12}+z^6}+\frac{z^2}{x^6+y^6+z^{12}}\le \frac{1}{x^2y^2z^2}. \)

[Claudiu Mindrila]

Notam \( x=a^2,y=b^2,z=c^2 \). Inegalitatea devine \( \sum \frac{a}{a^6+b^3+c^3} \leq\frac{1}{abc} \). Dar din inegalitatea mediilor
\( \sum \frac{a}{a^6+b^3+c^3} \leq \sum\frac{a}{3a^2bc}=\sum \frac{abc}{3a^2b^2c^2}=\frac{3abc}{3a^2b^2c^2}=\frac{1}{abc} \qed \)

[maxim bogdan]

Notam \( x=a^2,y=b^2,z=c^2 \).

Era \( x^2=a; y^2=b; z^2=c \)