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Jul 31, 2015 at 21:30 vote accept salimmath15
Jul 31, 2015 at 15:17 answer added zeraoulia rafik timeline score: 0
Jun 15, 2015 at 17:51 vote accept salimmath15
Jul 31, 2015 at 21:28
S Jun 15, 2015 at 17:48 history bounty ended CommunityBot
S Jun 15, 2015 at 17:48 history notice removed CommunityBot
Jun 8, 2015 at 15:33 answer added Iosif Pinelis timeline score: 3
S Jun 7, 2015 at 16:44 history bounty started salimmath15
S Jun 7, 2015 at 16:44 history notice added salimmath15 Draw attention
May 30, 2015 at 13:01 comment added GH from MO @rafik: Thanks, I deleted my answer now.
May 30, 2015 at 12:57 comment added salimmath15 @GHfromMO, i knew this that's published conjecture after positing it , in the fisrt it was my independent work
May 30, 2015 at 12:45 comment added salimmath15 @GHfromMO you can edit your answer
May 30, 2015 at 8:14 history edited YCor CC BY-SA 3.0
Changed to a useful title. Added reference from the comments.
May 30, 2015 at 3:43 history edited Arturo Magidin CC BY-SA 3.0
As long as there was a recent bump, let's get rid of all those nasty typos, the annoying "how to proof", etc.
May 30, 2015 at 2:39 review Close votes
May 30, 2015 at 11:47
May 30, 2015 at 1:47 comment added GH from MO As @RobertIsrael pointed out to me, the problem asked above is Conjecture 8 in the paper anubih.ba/Journals/vol.8,no-2,y12/11Ladas-Lugo-Palladino.pdf . My answer below was accepted by the OP, but unfortunately it answered a much simpler question (due to my misunderstanding of the original question). Ideally I would delete my answer, but I cannot, because it is protected now.
May 25, 2015 at 12:17 comment added salimmath15 @JosephO'Rourke, only what i have that this conjecture deal with the following simpler system: $$\begin{cases} x_{n+1}=\dfrac{\alpha_{1}}{y_{n}} \\ y_{n+1}=\dfrac{\alpha_{2}}{z_{n}} \\ z_{n+1}=\dfrac{\alpha_{3+}+\beta_{3}x_{n}+\sigma_{3}y_{n}+\lambda_{3}z_{n}}{A_{3}+B_{3}x_{n}+c_{3}y_{n}+D_{3}z_{n}}{}\end{cases} \quad n=0,1,\dots,$$ .with non-negative paramaters and non-negative initial conditions and dominators are never zero
May 25, 2015 at 10:24 vote accept salimmath15
May 30, 2015 at 12:52
May 24, 2015 at 22:35 comment added Joseph O'Rourke Despite your lack of interest in initial values, perhaps your conjecture is True for initial values of $1$.
May 24, 2015 at 21:49 comment added salimmath15 no, i'm not intended any specific initial values
May 24, 2015 at 19:12 comment added Joseph O'Rourke Perhaps you intended some specific initial values, e.g., $z_0=z_1=z_2=1$...?
May 24, 2015 at 18:15 history asked salimmath15 CC BY-SA 3.0