Inegalitate
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Inegalitate
Fie \( a,b,c,d\in \mathbb{R}_+ \) astfel incat : \( \frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{1}{1+c^4}+\frac{1}{1+d^4}=1 \) . Sa se arate ca \( abcd\ge 3 \) .
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maxim bogdan
- Thales
- Posts: 106
- Joined: Tue Aug 19, 2008 1:56 pm
- Location: Botosani
Relatia din enunt este echivalenta cu:
\( 3+2\sum a^4+\sum a^4b^4=a^4b^4c^4d^4, \) unde \( \sum a^4=a^4+b^4+c^4+d^4 \) si \( \sum a^4b^4=a^4b^4+a^4c^4+a^4d^4+b^4c^4+b^4d^4+c^4d^4. \)
Din inegalitatea mediilor vom avea:
\( \sum a^4\geq 4abcd \ \wedge \ \sum a^4b^4\geq 6a^2b^2c^2d^2. \)
Notam \( abcd=k. \) Folosind observatia de mai sus vom avea:
\( k^4\geq 6k^2+8k+3\Longleftrightarrow (k-3)(k^3+3k^2+3k+1)\geq 0\Longleftrightarrow (k-3)(k+1)^3\geq 0. \) Cum \( (k+1)^3>0 \) deducem ca \( k\geq 3 \) adica \( abcd\geq 3. \)
\( 3+2\sum a^4+\sum a^4b^4=a^4b^4c^4d^4, \) unde \( \sum a^4=a^4+b^4+c^4+d^4 \) si \( \sum a^4b^4=a^4b^4+a^4c^4+a^4d^4+b^4c^4+b^4d^4+c^4d^4. \)
Din inegalitatea mediilor vom avea:
\( \sum a^4\geq 4abcd \ \wedge \ \sum a^4b^4\geq 6a^2b^2c^2d^2. \)
Notam \( abcd=k. \) Folosind observatia de mai sus vom avea:
\( k^4\geq 6k^2+8k+3\Longleftrightarrow (k-3)(k^3+3k^2+3k+1)\geq 0\Longleftrightarrow (k-3)(k+1)^3\geq 0. \) Cum \( (k+1)^3>0 \) deducem ca \( k\geq 3 \) adica \( abcd\geq 3. \)
Feuerbach
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maxim bogdan
- Thales
- Posts: 106
- Joined: Tue Aug 19, 2008 1:56 pm
- Location: Botosani
Ca o observatie, asta este inegalitatea data la barajul din Letonia 2002. O alta solutie se bazeaza pe substitutia \( a^2=\tan A, \ A\in (0;\frac{\pi}{2}), \) si relatiile analoage. De aici se observa ca conditia din enunt e echivalenta cu:
\( \sum \cos^2 A=1. \) De aici e usor!
A se vedea urmatorul link.
\( \sum \cos^2 A=1. \) De aici e usor!
A se vedea urmatorul link.
Feuerbach