Inegalitate nice
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Claudiu Mindrila
- Fermat
- Posts: 520
- Joined: Mon Oct 01, 2007 2:25 pm
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Inegalitate nice
Demonstrati ca pentru orice \( x,y,z \in (0, \infty) \) are loc inegalitatea: \( \frac{x}{xy+x+1}+\frac{y}{yz+y+1}+\frac{z}{zx+x+1} \leq 1. \)
elev, clasa a X-a, C. N. "C-tin Carabella", Targoviste
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Marius Mainea
- Gauss
- Posts: 1077
- Joined: Mon May 26, 2008 2:12 pm
- Location: Gaesti (Dambovita)
Inegalitatea este echivalenta cu
\( \sum_{cyc} {\frac{1}{y+1+\frac{1}{x}}}\le1 \) sau
\( \sum_{cyc} {(\frac{1}{x}+y+1)(\frac{1}{y}+z+1)}\le (\frac{1}{x}+y+1)(\frac{1}{y}+z+1)(\frac{1}{z}+x+1) \) sau
\( \sum_{cyc} {[\frac{1}{xy}+\frac{z}{x}+\frac{2}{x}+2+yz+2y]}\le \frac{1}{xyz}+xyz+4+\sum_{cyc} {[\frac{2}{x}+\frac{1}{xy}+2x+\frac{z}{x}+xy]} \) sau
\( 2\le xyz+\frac{1}{xyz} \) care este evident adevarata , egalitatea avand loc daca si numai daca \( xyz=1 \)
\( \sum_{cyc} {\frac{1}{y+1+\frac{1}{x}}}\le1 \) sau
\( \sum_{cyc} {(\frac{1}{x}+y+1)(\frac{1}{y}+z+1)}\le (\frac{1}{x}+y+1)(\frac{1}{y}+z+1)(\frac{1}{z}+x+1) \) sau
\( \sum_{cyc} {[\frac{1}{xy}+\frac{z}{x}+\frac{2}{x}+2+yz+2y]}\le \frac{1}{xyz}+xyz+4+\sum_{cyc} {[\frac{2}{x}+\frac{1}{xy}+2x+\frac{z}{x}+xy]} \) sau
\( 2\le xyz+\frac{1}{xyz} \) care este evident adevarata , egalitatea avand loc daca si numai daca \( xyz=1 \)