Inegalitate exponentiala

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Marius Mainea
Gauss
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Inegalitate exponentiala

Post by Marius Mainea »

Sa consideram a,b,c>0 si x>y>0.
Atunci \( 2[(ab)^x+(bc)^x+(ca)^x]\ge a^{2x}+b^{2x}+c^{2x} \) implica

\( 2[(ab)^y+(bc)^y+(ca)^y]\ge a^{2y}+b^{2y}+c^{2y} \)

S. Radulescu, D. Popescu, Concursul Arhimede, 28-02-2009
smileyRadu
Arhimede
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Post by smileyRadu »

O solutie ar fi sa treci in totul in membrul stang in prima inegaliate,consideri functia si o derivezi. Derivata fiind pozitiva rezulta ca functia este crescatoare si cum x>y rezulta concluzia.
R.P.
Marius Mainea
Gauss
Posts: 1077
Joined: Mon May 26, 2008 2:12 pm
Location: Gaesti (Dambovita)

Post by Marius Mainea »

Si alta de clasa a -X-a ?
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