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I am blaming this on https://math.stackexchange.com/questions/891133/prove-the-equality

EDITTTTT: allowing $x_1 \geq x_2$ and $x_2 \geq x_n,$ I would rather not explain what that was about and the only changes are in $n=3,4,$ already settled.

Question: Given integers $x_1,x_2,x_3, \ldots, x_n \geq 1,$ with $x_1 \geq x_2,x_3, \ldots, x_n, $ and $$\color{magenta}{ x_1^2 \leq x_2^2 - x_2 x_3 + x_3^2 - x_3 x_4 + x_4^2 - \cdots + x_{n-1}^2 - x_{n-1} x_n + x_n^2}, $$ can we prove that $$ \color{blue}{ x_1^2 + x_2^2 + \cdots + x_{n-1}^2 + x_n^2 - x_1 x_2 - x_2 x_3- \cdots - x_{n-1} x_n - x_n x_1 < x_1 x_2 x_3 \cdots x_n}? $$

Pointed out by Thomas Andrews: $$ \color{green}{ x_1^2 + x_2^2 + \cdots + x_{n-1}^2 + x_n^2 - x_1 x_2 - x_2 x_3- \cdots - x_{n-1} x_n - x_n x_1 = \frac{1}{2} \left( (x_1 - x_2)^2 + (x_2 - x_3)^2 +\cdots + (x_n - x_1)^2 \right)}. $$

NOTES: for $n=3,$ the condition in magenta is pretty final, because $$ x_1^2 \leq x_2^2 - x_2 x_3 + x_3^2 $$ implies that $$ x_1 = x_2 = x_3. $$

I was also able to prove the final conclusion for $n=4.$ It appears, for $n=4,$ that $$ \color{blue}{ x_1^2 + x_2^2 + x_3^2 + x_4^2 - x_1 x_2 - x_2 x_3- x_3 x_4 - x_4 x_1 \leq \frac{4}{21} x_1 x_2 x_3 x_4} $$ with equality at $$ (7,3,1,7) \; \; \mbox{and} \; \; (7,7,1,3). $$ I think, for $n=4,$ that allowing the variables to take real values $\geq 1$ with $x_1$ maximal but equality allowed, the optimum occurs at $(T^2 - T + 1, T^2 - T + 1, 1, T),$ with $$ T = 1 + \sqrt[3]{1 + \sqrt{\frac{19}{27}}} + \sqrt[3]{1 - \sqrt{\frac{19}{27}}} \approx 2.769292354 $$ and $T^2 - T + 1 \approx 5.899687788.$ The apparent best ratio for integers, $4/21 \approx 0.190476,$ is improved to about $0.191602$ for real numbers.

It appears, for $n=5,$ that $$ \color{blue}{ x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 - x_1 x_2 - x_2 x_3- x_3 x_4 - x_4 x_5 - x_5 x_1 \leq \frac{1}{3} x_1 x_2 x_3 x_4 x_5} $$ with equality at $$ (4,1,4,1,3) \; \; \mbox{and} \; \; (4,3,1,4,1). $$ I think, for $n=5,$ that allowing the variables to take real values $\geq 1$ with $x_1$ maximal but equality allowed, the optimum occurs at $((U^2 - U + 2)/2, \; 1, \; (U^2 - U + 2)/2, \; 1, \; U),$ with $$ U = 1 + \sqrt[3]{2 + \sqrt 3} + \sqrt[3]{2 - \sqrt3} \approx 3.1958233 $$ and $ (U^2 - U + 2)/2 \approx 4.50873.$ The apparent best ratio for integers, $1/3 \approx 0.33333,$ is improved to about $0.33462442$ for real numbers.

It appears, for $n \geq 6,$ that $$ \color{blue}{ x_1^2 + x_2^2 + \cdots + x_{n-1}^2 + x_n^2 - x_1 x_2 - x_2 x_3- \cdots - x_{n-1} x_n - x_n x_1 \leq \frac{4}{9} x_1 x_2 x_3 \cdots x_n}, $$ with equality at $$ (3,1,1,\ldots,3,1,1,\ldots,3,1,\ldots,1) $$ which is to say that all but three entries are $1,$ while $x_1 = 3,$ then $x_i = x_j = 3$ with $3 \leq i,j \leq n-1$ and $i \leq j-2.$ i do not think anything changes by allowing real numbers when $n \geq 6.$

BACKGROUND: the strange condition $x_1^2 \leq \mbox{STUFF}$ describes what Hurwitz called a "fundamental solution" or Grundlösung, or Grundlösung for the original Diophantine equation. If anyone knows how to do an umlaut in text mode let me know (Thank you Hans Lundmark and miracle173). Most people on this site will be familiar with the phrase Vieta Jumping, which is part of what Hurwitz was doing (1907). Let's see, I believe I have abstracted out the number theory part to a great extent, a proof for what is left establishes the conjecture in the earlier question for all $n.$ That is, i think this stuff is true (with worse constants) for real numbers, all at least $1,$ the same strange quadratic inequality.

Note that, if we put no upper bound on $x_1,$ allowing it to grow without bound but keeping the others fixed, the left hand side is eventually larger than any fixed multiple of the right hand side, as the left is quadratic in $x_1$ while the right is linear in $x_1.$ The strange upper bound is the whole story.

BACK OF BACKGROUND. After the paper on the Markov Numbers appeared, Hurwitz decided to investigate $$ x_1^2 + x_2^2 + \cdots + x_n^2 = a x_1 x_2 \cdots x_n $$ with complete success. For each fixed $n$ and $a,$ there are a finite number of fundamental solutions defined by taking only the solutions that satisfy an inequality. For each such fundamental solution, there is an entire tree of solutions, and tree traversal from one solution to a neighbor or Nachbarn is by Vieta Jumping in a single one of the variables. A fundamental solution is one for which any immediate neighbor has all entries at least as large as the originals; he refers to minimizing the height or Höhe $x_1 + x_2 + \cdots + x_n.$ For $n \leq 13$ there is at most one Grundlösung for a given $(n,a),$ but there are two for $n=14, a=1.$ Fairly complete description and reference at https://mathoverflow.net/questions/84927/conjecture-on-markov-hurwitz-diophantine-equation

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    $\begingroup$ Here's an "ö" for you to cut & paste if you like. I made it by pressing the "ö" key on my Swedish keyboard. :-) $\endgroup$ – Hans Lundmark Aug 12 '14 at 5:07
  • $\begingroup$ @HansLundmark, thank you, that worked. I have no trouble doing this in actual Latex preprints, often my bibliography section has many umlauts; but that does not work in MathJax. $\endgroup$ – Will Jagy Aug 12 '14 at 18:15
  • $\begingroup$ One can use the html technic: "ö" is written as "&ouml;" or "&#246;". This works in questions and answers but not in comments. de.selfhtml.org/html/referenz/zeichen.htm#benannte_iso8859_1 $\endgroup$ – miracle173 Aug 12 '14 at 21:03
  • $\begingroup$ @miracle173, thanks, do we include the quotation marks? I see, no quotation marks but do include the semicolon. $\endgroup$ – Will Jagy Aug 12 '14 at 21:07
  • $\begingroup$ not quotation marks, but semicolon. On my Windows notebook I can use the Alt Numpad input method to produce "ö" bei pressing directly "[Alt]+148". [Alt] is the left Alt-key, 148 must be typed on the numeric keypad. This works in comments, too. These codes work for me: en.wikipedia.org/wiki/Code_page_437#Characters $\endgroup$ – miracle173 Aug 12 '14 at 21:18

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