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inegalitate

 
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seby97
Euclid


Joined: 04 Aug 2011
Posts: 31

PostPosted: Fri Aug 12, 2011 4:57 pm    Post subject: inegalitate Reply with quote

Sa se demonstreze ca daca a,b,c>0 si abc=1,atunci
1/a +1/b +1/c +6/a+b+c >=5
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smecherul100
Euclid


Joined: 09 Nov 2010
Posts: 37

PostPosted: Sat Mar 03, 2012 4:09 pm    Post subject: Reply with quote

plicam cauchy-schwartz pentru 1/a,/b,1/c si a,b,c si avem 1/a+1/b+1/c >3/(a+b+c)
,adica ar fi suficient sa dem ca 15/(a+b+c) >5 <3>1 <3>a+b+c,fals!!!!

de ex:a=3,b=1,c=1/3.cred ca conditia abc=1 nu este buna. Very Happy
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PhantomR
Pitagora


Joined: 20 Mar 2011
Posts: 86

PostPosted: Sat Mar 03, 2012 6:38 pm    Post subject: Reply with quote

Contraexemplul este invalid.
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smecherul100
Euclid


Joined: 09 Nov 2010
Posts: 37

PostPosted: Sun Mar 04, 2012 1:04 pm    Post subject: Reply with quote

dar totusi inegalitatea nu pare buna... Sad
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seby97
Euclid


Joined: 04 Aug 2011
Posts: 31

PostPosted: Wed Oct 10, 2012 3:18 pm    Post subject: Reply with quote

E corecta,putand fi rezolvata usor cu Mixing Variables!
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