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Fie \( x,\ y,\ z\in\mathbb{R}_{+}^{*} \) astfel incat \( x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}=3x^{2}y^{2}z^{2} \). Demonstrati ca \( \frac{1}{x^{2}+x+1}+\frac{1}{y^{2}+y+1}+\frac{1}{z^{2}+z+1}\le1 \).

Razvan Ceuca, Recreatii Matematice 1/2009

\( \frac{3x^2y^2z^2}{x^2y^2+y^2z^2+z^2x^2}=1 \Rightarrow \frac{3}{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}}=1 \Rightarrow \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=3. \)
Aplicand inegalitatea Cauchy-Buniakowski-Schwarz (forma Titu Andreescu) avem: \( \frac{(\frac{1}{x})^2}{1}+\frac{(\frac{1}{y})^2}{1}+\frac{(\frac{1}{z})^2}{1}\geq \frac{(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2}{(1+1+1)^2}, \) adica \( \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\geq \frac{(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2}{3}\Rightarrow \)
\( \Rightarrow (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2\leq 9 \Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\leq 3 \).
\( \frac{1}{x^2+x+1}\leq \frac{1}{3x} \) deoarece \( x^2+x+1\geq 3x; x^2-2x+1\geq 0; (x-1)^2\geq 0 \), evident.
Analog pentru \( \frac{1}{y^2+y+1} \) si \( \frac{1}{z^2+z+1} \).
Atunci \( \frac{1}{x^2+x+1}+\frac{1}{y^2+y+1}+\frac{1}{z^2+z+1}\leq \frac{1}{3}(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\leq \frac{1}{3}\cdot 3=1 \).

"Aplicand inegalitatea Cauchy-Buniakowski-Schwarz (forma Titu Andreescu)" !!!??!!
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L.O.