2024 CMO

(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 $$f(x)=(x-a)(x-b{)}^2=x^3-(a+2b)x^2+(2ab+b^2)x-a{b}^2,$$ and define $$S=\sum _{k=1}^{n}f(x_k)=3n-(a+2b)\cdot 2n+(2ab+b^2)\cdot n-a{b}^2\cdot n.$$ 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 $$\begin{array}{rl}\sum _{k=1}^n(x_k-A)& =\left(\sum _{k=1}^n{x}_k\right)-nA=n-nA,\\ \sum _{k=1}^n(x_k-A{)}^2& =2n-2nA+n{A}^2,\\ \sum _{k=1}^n(x_k-A{)}^3& =3n-6nA+3n{A}^2-n{A}^3,\end{array}$$ 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 $$z_1+z_2+\cdots +z_n=pn,z_1^2+z_2^2+\cdots +z_n^2=pn,z_1^3+z_2^3+\cdots +z_n^3=pn.$$ Let $\Vert x\Vert$ denote the distance of a real number $x$ to the nearest integer. It is well-known that $\Vert x+y\Vert \le \Vert x\Vert +\Vert y\Vert$ for any real numbers $x,y$. Let $m$ be the integer closest to $pn$, then $$\Vert pn\Vert =\left|m-\frac{\sqrt{5}+1}{2\sqrt{5}}n\right|=\frac{\left|(2m-n)\sqrt{5}-n\right|}{2\sqrt{5}}=\frac{4}{2\sqrt{5}}\cdot \frac{\left|5m^2-5mn+n^2\right|}{(2m-n)\sqrt{5}+n}\ge \frac{4}{2\sqrt{5}}\cdot \frac{1}{2n+1}.$$

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 $\delta >\frac{\lambda }{\sqrt{5n^{1.5}}}$. First, assume $\delta \le \frac{1}{2}$. Consider the polynomial $h(z)=1-3z^2+2z^3=(1+2z)(1-z{)}^2$. When $z\in [-\delta ,\frac{1}{2}]$, $\Vert z\Vert =|z|$, we have $$\Vert h(z)\Vert \le |1-h(z)|=(3-2z)z^2\le (3+2\delta )z^2\le 4z^2=4\Vert z{\Vert }^2.$$ When $z\in [\frac{1}{2},1+\delta ]$, $\Vert z\Vert =|z-1|$, we have $$\Vert h(z)\Vert \le |h(z)|=(1+2z)(1-z{)}^2\le (3+2\delta )(1-z{)}^2\le 4(1-z{)}^2=4\Vert z{\Vert }^2.$$ Thus, in all cases, $\Vert h(z_k)\Vert \le 4\Vert z_k{\Vert }^2$. Since ${\sum }_{k=1}^{n}h(z_k)=n-3\cdot pn+2\cdot pn=n-pn$, it follows that $$\sum _{k=1}^n\Vert h(z_k)\Vert \ge \Vert pn\Vert ,$$ and $$(\Vert z_1\Vert +\Vert z_2\Vert +\cdots +\Vert z_n\Vert {)}^2\ge \Vert z_1{\Vert }^2+\Vert z_2{\Vert }^2+\cdots +\Vert z_n{\Vert }^2\ge \frac{1}{4}\Vert pn\Vert .$$ On the other hand, consider the polynomial $g(z)=2z(1-z)$. When $z\in [0,1]$, we have $g(z)=2z(1-z)\ge \max \{z,1-z\}=\Vert z\Vert$. When $z\in [-\delta ,0]\cup [1,1+\delta ]$, we have $g(z)\ge -2(\delta +{\delta }^2)\ge \Vert z\Vert -4\delta$. Thus, $g(z_k)\le \Vert z_k\Vert -4\delta$. 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 $$4n\delta \ge \sum _{k=1}^n\Vert z_k\Vert \ge \sqrt{\frac{1}{4}\Vert pn\Vert }\ge \sqrt{\frac{1}{2\sqrt{5}}\cdot \frac{1}{2n+1}}>\sqrt{\frac{1}{11n}}.$$ 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. $\square$