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Fie x,y,z>0 cu produsul 1. Demonstrati ca :

\( 27(x^3+1)(y^3+1)(z^3+1)\ge (x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^3. \)

Avem urmatorul rezultat mai general.
Generalizare. Daca \( x_{1},x_{2},\ldots,x_{n}>0 \) astfel incat \( x_{1}x_{2}\cdot\ldots\cdot x_{n}=1 \) si \( n \in \mathbb{N} \), atunci are loc inegalitatea: \( n^{n}\left(x_{1}^{n}+1\right)\left(x_{2}^{n}+1\right)\cdot\ldots\cdot\left(x_{n}^{n}+1\right)\geq\left(x_{1}+x_{2}+\ldots+x_{n}+\frac{1}{x_{1}}+\frac{1}{x_{2}}+\ldots+\frac{1}{x_{n}}\right)^{n}. \)

Demonstratie. Conform inegalitatii Huygens avem: \( \prod_{i=1}^{n}\left(x_{i}+1\right)\geq\left(x_{1}+\prod_{i=2}^{n}x_{i}\right)^{n}=\left(x_{1}+\frac{1}{x_{1}}\right)^{n} \) de unde \( \sqrt[n]{\prod_{i=1}^{n}\left(x_{i}+1\right)}\geq x_{1}+\frac{1}{x_{1}} \) si chiar \( n\cdot\sqrt[n]{\prod_{i=1}^{n}\left(x_{i}+1\right)}\geq\sum_{1}^{n}x_{i}+\sum_{i=1}^{n}\frac{1}{x_{i}} \). Prin ridicare la puterea \( n \) a ultimei inegalitati rezulta ca \( n^{n}\cdot\prod_{i=1}^{n}\left(x_{i}+1\right)\geq\left(\sum_{i=1}^{n}x_{i}+\sum_{i=1}^{n}\frac{1}{x_{i}}\right)^{n} \) ceea ce trebuia aratat.

Caz particular. Pentru \( n=3 \) se obtine inegalitatea ceruta.