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An equivalent inequality for integers follows:

$$(3n^2-5)\left\lceil n/\sqrt{3} \right\rceil^2 > n^4.$$

This has been checked for n = 2 to 60000. Perhaps there is some connection to the convergents to $\sqrt{3}$.

$\lceil \frac{n}{\sqrt{3}} \rceil > \frac{n^2}{\sqrt{3n^2-5}}$

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    $\begingroup$ Should follow from $|a/b -\sqrt{3}| > \alpha/b^2$, which is true for all relatively prime $a,b$ for a suitable $\alpha$. $\endgroup$ – Felipe Voloch Nov 6 '14 at 22:59
  • $\begingroup$ If you like my answer, please accept it officially (so that it turns green). Thanks in advance! $\endgroup$ – GH from MO Aug 19 '18 at 23:22
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The inequality is true for $n\geq 2$. Let $m:=\lceil n/\sqrt{3}\rceil$, then $$ m^2-\frac{n^2}{3} = \frac{3m^2-n^2}{3} \geq \frac{2}{3}, $$ because $3m^2-n^2>0$ and $3m^2-n^2\not\equiv 1\pmod{3}$. Hence the inequality follows from $$ \frac{2}{3}>\frac{n^4}{3n^2-5}-\frac{n^2}{3}=\frac{5n^2}{3(3n^2-5)}. $$ The latter is equivalent to $n^2>10$, hence we have a proof for $n\geq 4$. The remaining cases $n=2$ and $n=3$ can be checked by hand.

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  • $\begingroup$ Why is $\frac{3m^2-n^2}{3}\geq \frac{2}{3}$? $\endgroup$ – Alan Nov 7 '14 at 7:00
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    $\begingroup$ @Alan, the left side is something-nonnegative-over-3, and GH tells you why that something isn't zero or one, so it's at least 2. $\endgroup$ – Gerry Myerson Nov 7 '14 at 8:58
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Equivalently, we want to know if

$\mathrm{sqceiling}(n^2/3) > n^4/(3n^2-5) = n^2/3 + 5/9 + 25/(27n^2) + ...$

where $\mathrm{sqceiling}()$ is the function taking a real to the next exact square.

This is equivalent to

$\mathrm{sqceiling}(n^2/3) - n^2/3 > 5/9 + 25/(27n^2) + ...$

but since $n^2/3$ is never a square for $n>0$, and $n^2 \equiv 0\ \textrm{or}\ 1 (\textrm{mod}\ 3)$, we have

$\mathrm{sqceiling}(n^2/3) - n^2/3 \ge 2/3$.

Clearly for large enough $n$

2/3 > 5/9 + 25/(27n^2) + ...

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  • $\begingroup$ We wrote much the same, but you beat me by 4 seconds! $\endgroup$ – GH from MO Nov 6 '14 at 23:15
  • $\begingroup$ not quite! I was still editing my latex. $\endgroup$ – Yaakov Baruch Nov 6 '14 at 23:21
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A tight result is $$\lceil \frac{n}{\sqrt{3}}\rceil \ge \frac{n^2}{\sqrt{3n^2-6+\frac{12}{n^2+2}}}$$

This means that the desired inequality holds for $n \ge 4.$ The cases $n=2,3$ can be checked separately.

As noted elsewhere, for $m=\lceil \frac{n}{\sqrt{3}}\rceil$ we have $n^2 \le 3m^2-2$ with equality exactly for the cases $n=1,5,19,71,265,989,\cdots$ A001834 with $\frac{n}{m} \lt \sqrt{3}$ a convergent to that square root. A bit of manipulation then gives the result above and the information that equality holds for exactly those values of $n.$

Details: Change the desired inequality to $$m^2 \ge \frac{n^4}{3n^2-k}$$ with $k$ to be determined. Since $n^2=3m^2-j$ with $j \ge 2,$ we have $m^2=\frac{n^2+j}{3}.$ This yields $$k \le \frac{3jn^2}{n^2+j} =3j-\frac{3j^2}{n^2+j}.$$ Since we do sometime have $j=2,$ The bound $k \le 6-\frac{12}{n^2+2}$ is sometime an equality. It is not hard to show that $\frac{3jn^2}{n^2+j}$ increases when $j \ge 2$ does.

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