, 2024

(1)

Solution 1
We attempt to use the idea of a local inequality to prove that a number $a$ lies between the minimum and maximum of a regular $n$-tuple. To this end, we consider an indeterminate $b$ and consider the polynomial
$$f(x) = (x - a)(x - b)^2 = x^3 - (a + 2b)x^2 + (2ab + b^2)x - ab^2,$$
and define
$$S = \sum_{k=1}^{n} f(x_k) = 3n - (a + 2b) \cdot 2n + (2ab + b^2) \cdot n - ab^2 \cdot n.$$
We require $S = 0$ to ensure that $f(x_1), f(x_2), \dots, f(x_n)$ have both positive and negative values (or are all 0), implying that the numbers $x_1, x_2, \dots, x_n$ include values $\ge a$ and values $\le a$ (or $x_1, \dots, x_n$ are all equal to $b$). This ensures $a$ lies between the minimum and maximum values.
Setting $S = 0$, we obtain the quadratic equation
$$(1 - a)b^2 + (2a - 4)b + (3 - 2a) = 0.$$
This equation has real roots if and only if
$$\Delta = (2a - 4)^2 - 4(1 - a)(3 - 2a) = 4(-a^2 + a + 1) \ge 0,$$
which is equivalent to $\frac{1-\sqrt{5}}{2} \le a \le \frac{1+\sqrt{5}}{2}$. Thus, the maximum value $\ge \frac{1+\sqrt{5}}{2}$ and the minimum value $\le \frac{1-\sqrt{5}}{2}$. The width is therefore $\ge \frac{1+\sqrt{5}}{2} - \frac{1-\sqrt{5}}{2} = \sqrt{5}$.
On the other hand, we can construct a regular $n$-tuple whose width approaches $\sqrt{5}$ arbitrarily closely.
Take positive integers $u_1, v_1, u_2, v_2$ such that $\frac{v_1}{u_1} > \frac{\sqrt{5}-1}{2} > \frac{v_2}{u_2}$.
Define the first group (multiset) to include $u_1^2$ instances of $1 + \frac{v_1}{u_1}$ and $v_1^2$ instances of $1 - \frac{u_1}{v_1}$. The mean of these $u_1^2 + v_1^2$ numbers is 1, the mean square is 2, and the mean cube is $4 + \left( \frac{v_1}{u_1} - \frac{u_1}{v_1} \right) > 3$.

Define the second group (multiset) to include $u_2^2$ instances of $1 + \frac{v_2}{u_2}$ and $v_2^2$ instances of $1 - \frac{u_2}{v_2}$. The mean of these $u_2^2 + v_2^2$ numbers is 1, the mean square is 2, and the mean cube is $4 + \left(\frac{v_2}{u_2} - \frac{u_2}{v_2}\right) < 3$.
By combining sufficient numbers of the first and second groups, the overall mean cube can be made exactly 3, and the width $(1 + \frac{v_1}{u_1}) - (1 - \frac{u_2}{v_2}) = \frac{v_1}{u_1} + \frac{u_2}{v_2}$ can approach $\sqrt{5}$ arbitrarily closely.
Hence, the largest possible real number $C = \sqrt{5}$. $\square$

Solution 2
(The construction of regular $n$-tuples is the same as in Solution 1.)
Let $A$ and $B$ denote the minimum and maximum values among $x_1, x_2, \dots, x_n$, respectively.
For each $x_k - A \ge 0$, applying the Cauchy–Schwarz inequality, we have
$$\sum_{k=1}^{n} (x_k - A) \cdot \sum_{k=1}^{n} (x_k - A)^3 \ge \left( \sum_{k=1}^{n} (x_k - A)^2 \right)^2 .$$
Substituting
$$\begin{align*} \sum_{k=1}^{n} (x_k - A) &= \left(\sum_{k=1}^{n} x_k\right) - nA = n - nA, \\ \sum_{k=1}^{n} (x_k - A)^2 &= 2n - 2nA + nA^2, \\ \sum_{k=1}^{n} (x_k - A)^3 &= 3n - 6nA + 3nA^2 - nA^3, \end{align*}$$
we find that
$$(1 - A)(3 - 6A + 3A^2 - A^3) - (2 - 2A + A^2)^2 = A^2 - A - 1 \ge 0.$$
Solving, we obtain $A \le \frac{1-\sqrt{5}}{2}$.
Similarly, using the same inequality for $B - x_k \ge 0$, we find that $B \ge \frac{1+\sqrt{5}}{2}$.
Thus, the width $B - A \ge \sqrt{5}$. If we have an equality, then $A = \frac{1-\sqrt{5}}{2}$, $B = \frac{1+\sqrt{5}}{2}$, and all $x_1, \dots, x_n$ are either $A$ or $B$. This requires $\frac{5-\sqrt{5}}{10}n$ numbers equal to $A$ and $\frac{5+\sqrt{5}}{10}n$ numbers equal to $B$, which cannot hold if $n$ is not divisible by 10. Therefore, the width $B - A > \sqrt{5}$. $\square$

(2)

Consider a linear transformation that maps the approximate minimum value $\frac{1-\sqrt{5}}{2}$ to 0 and the approximate maximum value $\frac{1+\sqrt{5}}{2}$ to 1. Let $z_i = \frac{x_i - \frac{1-\sqrt{5}}{2}}{\sqrt{5}}$, i.e., $x_i = \frac{1-\sqrt{5}}{2} + \sqrt{5}z_i$. Denote $p = \frac{5+\sqrt{5}}{10} = \frac{\sqrt{5}+1}{2\sqrt{5}}$. Then, we have
$$z_1 + z_2 + \dots + z_n = pn, \quad z_1^2 + z_2^2 + \dots + z_n^2 = pn, \quad z_1^3 + z_2^3 + \dots + z_n^3 = pn.$$
Let $\|x\|$ denote the distance of a real number $x$ to the nearest integer. It is well-known that $\|x + y\| \le \|x\| + \|y\|$ for any real numbers $x, y$. Let $m$ be the integer closest to $pn$, then
$$\|pn\| = \left| m - \frac{\sqrt{5}+1}{2\sqrt{5}}n \right| = \frac{\left| (2m-n)\sqrt{5} - n \right|}{2\sqrt{5}} = \frac{4}{2\sqrt{5}} \cdot \frac{\left| 5m^2 - 5mn + n^2 \right|}{(2m-n)\sqrt{5} + n} \ge \frac{4}{2\sqrt{5}} \cdot \frac{1}{2n+1}.$$

From (1), we know that the maximum of $z_1, z_2, \dots, z_n$ is $\ge 1$, and the minimum is $\le 0$. Let the difference between the maximum and minimum values be $1 + \delta$, so $z_1, z_2, \dots, z_n \in [-\delta, 1 + \delta]$. We aim to prove that $\delta > \frac{\lambda}{\sqrt{5 n^{1.5}}}$. First, assume $\delta \le \frac{1}{2}$.
Consider the polynomial $h(z) = 1 - 3z^2 + 2z^3 = (1 + 2z)(1 - z)^2$.
When $z \in [-\delta, \frac{1}{2}]$, $\|z\| = |z|$, we have
$$\|h(z)\| \le |1 - h(z)| = (3 - 2z)z^2 \le (3 + 2\delta)z^2 \le 4z^2 = 4\|z\|^2.$$
When $z \in [\frac{1}{2}, 1 + \delta]$, $\|z\| = |z - 1|$, we have
$$\|h(z)\| \le |h(z)| = (1 + 2z)(1 - z)^2 \le (3 + 2\delta)(1 - z)^2 \le 4(1 - z)^2 = 4\|z\|^2.$$
Thus, in all cases, $\|h(z_k)\| \le 4\|z_k\|^2$.
Since $\sum_{k=1}^n h(z_k) = n - 3 \cdot pn + 2 \cdot pn = n - pn$, it follows that
$$\sum_{k=1}^{n} \|h(z_k)\| \ge \|pn\|,$$
and
$$(\|z_1\| + \|z_2\| + \cdots + \|z_n\|)^2 \ge \|z_1\|^2 + \|z_2\|^2 + \cdots + \|z_n\|^2 \ge \frac{1}{4} \|pn\|.$$
On the other hand, consider the polynomial $g(z) = 2z(1 - z)$.
When $z \in [0, 1]$, we have $g(z) = 2z(1 - z) \ge \max\{z, 1 - z\} = \|z\|$.
When $z \in [-\delta, 0] \cup [1, 1 + \delta]$, we have $g(z) \ge -2(\delta + \delta^2) \ge \|z\| - 4\delta$.
Thus, $g(z_k) \le \|z_k\| - 4\delta$.
Since $\sum_{k=1}^n g(z_k) = 2 \sum_{k=1}^n z_k - 2 \sum_{k=1}^n z_k^2 = 0$, we have
$$4n\delta \ge \sum_{k=1}^{n} \|z_k\| \ge \sqrt{\frac{1}{4} \|pn\|} \ge \sqrt{\frac{1}{2\sqrt{5}} \cdot \frac{1}{2n+1}} > \sqrt{\frac{1}{11n}}.$$
Thus, $\delta > \frac{1}{4\sqrt{11}} \cdot \frac{1}{n^{1.5}} > \frac{1}{6\sqrt{5n^{1.5}}}.$

Hence, the width of the regular $n$-tuple $(x_1, x_2, \dots, x_n)$ satisfies $> \sqrt{5} + \frac{1}{6n^{1.5}}$, and $\lambda = \frac{1}{6}$ meets the requirement. $\square$