Photaesthesia wrote: Let $a_1, a_2, \ldots, a_n$ be real numbers such that $\sum_{i=1}^n a_i = n$, $\sum_{i = 1}^n a_i^2 = 2n$, $\sum_{i=1}^n a_i^3 = 3n$. (i) Find the largest constant $C$, such that for all $n \geqslant 4$, \[ \max \left\{ a_1, a_2, \ldots, a_n \right\} - \max \left\{ a_1, a_2, \ldots, a_n \right\} \geqslant C. \](ii) Prove that there exists a constant $C_2$, such that \[ \max \left\{ a_1, a_2, \ldots, a_n \right\} - \max \left\{ a_1, a_2, \ldots, a_n \right\} \geqslant C + C_2 n^{-\frac 32}, \]where $C$ is the constant determined in (i). Did you mean max - min?

A rough solution ,may exist errors. **First Question:** Assume \( s = \max\{a_i\} \geq 1 \), \( t = \min\{a_i\} \leq 1 \), then \( s - a_i \) and \( a_i - t \) are both non-negative. By the Cauchy-Schwarz inequality, we have: \[ \sum (s - a_i) \cdot \sum (s - a_i)^3 \geq (\sum (s - a_i)^2)^2 \]and \[ \sum (a_i - t) \cdot \sum (a_i - t)^3 \geq (\sum (a_i - t)^2)^2. \]From these, we deduce \( s \geq \frac{1 + \sqrt{5}}{2} = 1.618\ldots \) and \( t \leq \frac{1 - \sqrt{5}}{2} = -0.618\ldots \), which implies \( s - t \geq \sqrt{5} \). When \( a_i \) are all equal to \( s \) and \( t \), and their quantities are in the ratio \( \sqrt{5} + 1 : \sqrt{5} - 1 = 1.618\ldots : 0.618\ldots \), the equality holds. However, since this ratio involves irrational numbers, the equality cannot be exactly achieved. For sufficiently large \( n \), the ratio can be arbitrarily approximated, and thus the maximum \( C \) is \( \sqrt{5} \). **Second Question:** Let \( u = \frac{1 + \sqrt{5}}{2} \) and \( v = \frac{1 - \sqrt{5}}{2} \). We can assume that \( v - 1 \leq a_i \leq u + 1 \) holds. Now consider the sum of \( f(a_i) = 2a_i^3 - 3a_i^2 - 6a_i \), where \( f'(u) = f'(v) = 0 \). Rewrite each \( a_i \) as \( u + b_i \) or \( v - b_i \), depending on which of \( u \) or \( v \) it is closer to. Assume there are \( p \) terms equal to \( u \) and \( n - p \) terms equal to \( v \). Then, \[ p f(u) + (n - p) f(v) - \sum f(a_i) \]is a rational approximation of an irrational number of the type \( \sqrt{5} \), which for sufficiently large \( n \) is at least \( \frac{C_3}{n} \). Moreover, \[ |f(u + b_i) - f(u)|, \, |f(v - b_i) - f(v)| \leq C_4 b_i^2, \]which implies \[ \sum b_i^2 \geq \frac{C_5}{n}. \]Thus, the sum of the squares of the positive \( b_i \) terms is at least \( \frac{C_6}{n} \), or the sum of the squares of the negative \( b_i \) terms is at least \( \frac{C_6}{n} \). If the sum of the squares of the positive \( b_i \) terms is at least \( \frac{C_6}{n} \), then the square of the largest positive term is at least \( \frac{C_6}{n^2} \), which implies that this term is at least \( \frac{C_7}{n} \). Consequently, there exists at least one \( a_i \) such that \( a_i \geq u + \frac{C_7}{n} \) or \( a_i \leq v - \frac{C_7}{n} \), and thus \( \max - \min \geq C + \frac{C_7}{n} \). If the sum of the squares of the negative \( b_i \) terms is at least \( \frac{C_6}{n} \), then the total absolute value of the negative terms is at least \( \frac{C_8}{\sqrt{n}} \). In this case, \[ \sum (u - a_i)(a_i - v) = 0, \]and \( (u - a_i)(a_i - v) \) is at least a constant multiple of \( |b_i| \). This implies that the sum of the positive terms is at least \( \frac{C_9}{\sqrt{n}} \), and thus the sum of the negative terms is at most \( -\frac{C_9}{\sqrt{n}} \). The largest absolute value of these terms is at least \( C_{10} n^{-3/2} \), which implies that this term is at least \( u + C_{11} n^{-3/2} \) or \( v - C_{11} n^{-3/2} \). Consequently, \( \max - \min \geq C + C_{11} n^{-3/2} \).

trivial by tangent line trick??? Let $\alpha=\frac{1+\sqrt5}2$ and $\beta=\frac{1-\sqrt5}2$. Then, $\alpha+\beta=1$ and $\alpha\beta=-1$, and $\alpha^2-\alpha-1=\beta^2-\beta-1=0$. We have \begin{align*} \sum_{i=1}^n(a_i-\alpha)^2(a_i-\beta)&=\sum_{i=1}^n(a_i^2-2a_i\alpha+\alpha^2)(a_i-\beta)\\ &=\sum_{i=1}^na_i^3-a_i^2(2\alpha+\beta)+a_i(\alpha^2+2\alpha\beta)-\alpha^2\beta\\ &=3n-2n(\alpha+1)+n(\alpha-1)+\alpha n\\ &=0 \end{align*}and \begin{align*} \sum_{i=1}^n(a_i-\beta)^2(a_i-\alpha)&=\sum_{i=1}^n(a_i^2-2a_i\beta+\beta^2)(a_i-\alpha)\\ &=\sum_{i=1}^na_i^3-a_i^2(2\beta+\alpha)+a_i(\beta^2+2\beta\alpha)-\beta^2\alpha\\ &=3n-2n(\beta+1)+n(\beta-1)+\beta n\\ &=0. \end{align*} Therefore, since $(a_i-\alpha)^2(a_i-\beta)\geq0$ for all $a_i>\beta$, with equality if and only if $a_i=\alpha$, we get that if $\min\{a_1,a_2,\ldots,a_n\}>\beta$, then $a_i=\alpha$ for all $i$, which contradicts $\sum_{i=1}^na_i=n$. This means $\min\{a_1,a_2,\ldots,a_n\}\leq\beta$. Similarly, since $(a_i-\beta)^2(a_i-\alpha)\leq0$ for all $a_i<\alpha$, with equality if and only if $a_i=\beta$, we get that if $\max\{a_1,a_2,\ldots,a_n\}<\alpha$, then $a_i=\beta$ for all $i$, which also contradicts $\sum_{i=1}^na_i=n$. This means $\max\{a_1,a_2,\ldots,a_n\}\geq\alpha$, so since $\alpha-\beta=\sqrt5$, we get $C\geq\sqrt5$. We show $\sqrt5$ is the largest possible constant. Define the Fibonacci sequence $F$ such that $F_1=F_2=1$ and $F_i=F_{i-1}+F_{i-2}$ for $i\geq3$. Now, let $m$ be even. Consider the construction $$a_i=\begin{cases}\frac{F_{m-2}}{F_m}&1\leq i\leq F_m^2F_{m+1}\\\frac{F_{m+2}}{F_m}&F_m^2F_{m+1}+1\leq i\leq F_m^2(F_{m-1}+F_{m+1})\\1&F_m^2(F_{m-1}+F_{m+1})+1\leq i\leq F_{m-1}F_{m+1}(F_{m-1}+F_{m+1})\end{cases}$$ which is well defined since $F_{m-1}F_{m+1}-F_m^2=(-1)^m=1\geq0$, and $n=F_{m-1}F_{m+1}(F_{m-1}+F_{m+1})$. Now, we have \begin{align*} \sum_{i=1}^na_i&=F_m^2F_{m+1}\frac{F_{m-2}}{F_m}+F_m^2F_{m-1}\frac{F_{m+2}}{F_m}+(F_{m-1}+F_{m+1})(F_{m-1}F_{m+1}-F_m^2)1\\ &=F_mF_{m+1}(F_m-F_{m-1})+F_mF_{m-1}(F_m+F_{m+1})+(F_{m-1}+F_{m+1})(F_{m-1}F_{m+1}-F_m^2)\\ &=F_m^2F_{m+1}+F_m^2F_{m-1}+(F_{m-1}+F_{m+1})(F_{m-1}F_{m+1}-F_m^2)\\ &=F_{m-1}F_{m+1}(F_{m-1}+F_{m+1})\\ &=n\\ \sum_{i=1}^na_i^2&=F_m^2F_{m+1}\frac{F_{m-2}^2}{F_m^2}+F_m^2F_{m-1}\frac{F_{m+2}^2}{F_m^2}+(F_{m-1}+F_{m+1})(F_{m-1}F_{m+1}-F_m^2)1^2\\ &=F_{m+1}(F_m-F_{m-1})^2+F_{m-1}(F_m+F_{m+1})^2+(F_{m-1}+F_{m+1})(F_{m-1}F_{m+1}-F_m^2)\\ &=F_{m-1}F_m^2+F_{m+1}F_m^2+F_{m-1}F_{m+1}^2+F_{m-1}^2F_{m+1}+(F_{m-1}+F_{m+1})(F_{m-1}F_{m+1}-F_m^2)\\ &=2F_{m-1}F_{m+1}(F_{m-1}+F_{m+1})\\ &=2n\\ \sum_{i=1}^na_i^3&=F_m^2F_{m+1}\frac{F_{m-2}^3}{F_m^3}+F_m^2F_{m-1}\frac{F_{m+2}^3}{F_m^3}+(F_{m-1}+F_{m+1})(F_{m-1}F_{m+1}-F_m^2)1^3\\ &=\frac{F_{m+1}(F_m-F_{m-1})^3+F_{m-1}(F_m+F_{m+1})^3}{F_m}+(F_{m-1}+F_{m+1})(F_{m-1}F_{m+1}-F_m^2)\\ &=F_m^2(F_{m-1}+F_{m+1})+\frac{F_{m+1}(3F_mF_{m-1}^2-F_{m-1}^3)+F_{m-1}(3F_mF_{m+1}^2+F_{m+1}^3)}{F_m}+(F_{m-1}+F_{m+1})(F_{m-1}F_{m+1}-F_m^2)\\ &=\frac{F_{m+1}(3F_mF_{m-1}^2-F_{m-1}^3)+F_{m-1}(3F_mF_{m+1}^2+F_{m+1}^3)}{F_m}+F_{m-1}F_{m+1}(F_{m-1}+F_{m+1})\\ &=3F_{m-1}F_{m+1}(F_{m-1}+F_{m+1})-\frac{F_{m+1}F_{m-1}(-F_{m-1}^2+F_{m+1}^2)}{F_m}+F_{m-1}F_{m+1}(F_{m-1}+F_{m+1})\\ &=4F_{m-1}F_{m+1}(F_{m-1}+F_{m+1})-F_{m+1}F_{m-1}(F_{m-1}+F_{m+1})\\ &=3n. \end{align*} Then, $\max\{a_1,a_2,\ldots,a_n\}-\min\{a_1,a_2,\ldots,a_n\}=\frac{F_{m+2}-F_{m-2}}{F_m}$, and as $m$ approaches infinity, this value approaches $\alpha^2-\alpha^{-2}=\sqrt5$. Therefore, $\boxed{C=\sqrt5}$. For part two, $C_2=0$ works, so assume $C_2$ is positive. Now, we will show either $\max\{a_1,a_2,\ldots,a_n\}\geq\alpha+O(n^{-1.5})$ or $\min\{a_1,a_2,\ldots,a_n\}\leq\beta-O(n^{-1.5})$. Suppose this is not true. Recall that \begin{align*} \sum_{i=1}^n(a_i-\alpha)^2(a_i-\beta)=0\\ \sum_{i=1}^n(a_i-\beta)^2(a_i-\alpha)=0. \end{align*} If $a_1\geq1$ and $a_1\leq\alpha-\varepsilon$, then $\sum_{i=2}^n(a_i-\alpha)^2(a_i-\beta)=-(a_1-\alpha)^2(a_1-\beta)\leq-0.1\varepsilon^2$, so there exists $a_i$ such that $(a_i-\alpha)^2(a_i-\beta)\leq-0.1\varepsilon^2n^{-1}$, which implies $a_i\leq\beta-O(\varepsilon^2n^{-1})$, so $\varepsilon=o(n^{-0.25})$. Similarly, if $a_1\leq1$ and $a_1\geq\beta+\varepsilon$, then $\sum_{i=2}^n(a_i-\beta)^2(a_i-\alpha)=(a_1-\beta)^2(a_1-\alpha)\geq0.1\varepsilon^2$, so for some $a_i$, we get $(a_i-\beta)^2(a_i-\alpha)\geq0.1\varepsilon^2n^{-1}$, so $a_i\leq\alpha+O(\varepsilon^2n^{-1})$ implies $\varepsilon=o(n^{-0.25})$. Therefore, for all $i$, either $a_i\geq\alpha+o(n^{-0.25})$ or $a_i\leq\beta+o(n^{-0.25})$. Now, split the $a_i$ into two sequences $\alpha+\varepsilon_1$, $\alpha+\varepsilon_2$, $\ldots$, $\alpha+\varepsilon_a$ and $\beta-\delta_1$, $\beta-\delta_2$, $\ldots$, $\beta-\delta_b$ such that $a+b=n$ and $-o(n^{-0.25})<\varepsilon_i,\delta_i<o(n^{-1.5})$. Substituting into the equations gives $$\sum_{i=1}^a\varepsilon_i^2(\sqrt5+\varepsilon_i)-\sum_{i=1}^b\delta_i(5+2\sqrt5\delta_i+\delta_i^2)=0$$and $$\sum_{i=1}^a\varepsilon_i(5+2\sqrt5\varepsilon_i+\varepsilon_i^2)+\sum_{i=1}^b\delta_i^2(-\sqrt5-\delta_i)=0.$$This gives $$\sum_{i=1}^a\varepsilon_i^2(\sqrt5+\varepsilon_i)=\sum_{i=1}^b\delta_i(5+2\sqrt5\delta_i+\delta_i^2)$$and $$\sum_{i=1}^a\varepsilon_i(5+2\sqrt5\varepsilon_i+\varepsilon_i^2)=\sum_{i=1}^b\delta_i^2(\sqrt5+\delta_i).$$ From $\sum_{i=1}^na_i=n$, we get $$\sum_{i=1}^a\varepsilon_i-\sum_{i=1}^b\delta_i=a+b-a\alpha-b\beta=a+(b-a)\alpha=\frac{3ab-a^2-b^2}{(a-b)\alpha+b}.$$The right hand side has absolute value at least $O(n^{-1})$. These three equations will be the only equations we will use after this point. Since all of these equations are symmetric in $\varepsilon$ and $\delta$, we can assume without loss of generality $\left|\sum_{i=1}^a\varepsilon_i\right|\geq O(n^{-1})$. We get $$\sum_{i=1}^a5\varepsilon_i=\sum_{i=1}^b\delta_i^2(\sqrt5+o(n^{-0.25}))-\sum_{i=1}^a(2\sqrt5+o(n^{-0.25}))\varepsilon_i^2.$$The magnitude of the left hand side is at least $O(n^{-1})$, so $$\sum_{i=1}^a\varepsilon_i^2+\sum_{i=1}^b\delta_i^2\geq O(n^{-1}).$$Since the terms with $\varepsilon_i,\delta_i>0$ have $\varepsilon_i^2,\delta_i^2=o(n^{-3})$, the sum of the squares of the negative $\varepsilon_i$ and $\delta_i$ is also at least $O(n^{-1})$. Therefore, the sum of the magnitudes of the negative $\varepsilon_i$ and $\delta_i$ is at least $O(n^{-0.5})$. We also have $$\sum_{i=1}^a\varepsilon_i(5+(2\sqrt5+o(n^{-0.25}))\varepsilon_i)=\sum_{i=1}^b\delta_i^2(\sqrt5+o(n^{-0.25})),$$so $$5\sum_{i=1}^a\varepsilon_i+(2\sqrt5+o(n^{-0.25}))\sum_{i=1}^a\varepsilon_i^2=\sum_{i=1}^b\delta_i^2(\sqrt5+o(n^{-0.25})).$$If $\sum_{i=1}^b\delta_i^2\geq O(n^{-0.5})$, then $\sum_{i=1}^a\varepsilon_i(5+(2\sqrt5+o(n^{-0.25}))\varepsilon_i)\geq O(n^{-0.5})$. The left hand side is only positive when $\varepsilon_i\geq0$, so then $\varepsilon_i\geq O(n^{-1.5})$ for some $\varepsilon_i$, contradiction. Therefore, $\sum_{i=1}^b\delta_i^2=o(n^{-0.5})$. Similarly, $\sum_{i=1}^a\varepsilon_i^2=o(n^{-0.5})$. Adding $$5\sum_{i=1}^a\varepsilon_i+(2\sqrt5+o(n^{-0.25}))\sum_{i=1}^a\varepsilon_i^2=\sum_{i=1}^b\delta_i^2(\sqrt5+o(n^{-0.25}))$$and $$5\sum_{i=1}^b\delta_i+(2\sqrt5+o(n^{-0.25}))\sum_{i=1}^b\delta_i^2=\sum_{i=1}^a\varepsilon_i^2(\sqrt5+o(n^{-0.25})).$$Adding gives $$5\sum_{i=1}^a\varepsilon_i+5\sum_{i=1}^b\delta_i=\sum_{i=1}^a\varepsilon_i^2(-\sqrt5+o(n^{-0.25}))+\sum_{i=1}^b\delta_i^2(-\sqrt5+o(n^{-0.25}))=o(n^{-0.5}).$$ However, since the sum of the magnitudes of the negative $\varepsilon_i$ and $\delta_i$ is at least $O(n^{-0.5})$, we know that the sum of the magnitudes of the positive $\varepsilon_i$ and $\delta_i$ is also at least $O(n^{-0.5})$, so at least one $\varepsilon_i$ or $\delta_i$ is at least $O(n^{-1.5})$, giving the bound $\max\{a_1,a_2,\ldots,a_n\}-\min\{a_1,a_2,\ldots,a_n\}\geq\sqrt5+O(n^{-1.5})$.

Photaesthesia wrote: ………… (ⅱ) Prove that there exists a constant $C_2$, such that for all $n\geqslant4$, $$\max{\left\{a_1,a_2,\dotsc,a_n\right\}}- \min{\left\{a_1,a_2,\dotsc,a_n\right\}}\geqslant C+C_2n^{-\frac32}\text,$$where $C$ is the constant determined in (ⅰ). The original problem requires \(C_2>0\).

Rhapsodies_pro wrote: The original problem requires \(C_2>0\). Thanks for your complement.

(what I believe is the) official wording: For an integer $n \geq 4$, if the real numbers $x_1$, $x_2$, $\dots$, $x_n$ satisfy the following three equalities: \begin{align*} x_1 + x_2 + \dots + x_n &= n, \\ x_1^2 + x_2^2 + \dots + x_n^2 &= 2n, \\ x_1^3 + x_2^3 + \dots + x_n^3 &= 3n, \end{align*}then we call $(x_1, x_2, \dots, x_n)$ a good $n$-tuple, and we let the width of this $n$-tuple be the difference between the largest number and the smallest number in $x_1$, $x_2$, $\dots$, $x_n$. (1) Find the largest constant $C$ such that for all $n \geq 4$, the width of every good $n$-tuple is greater than or equal to $C$. (2) For the constant $C$ found in (1), prove that there exists a positive constant $\lambda$ such that for all $n \geq 4$, the width of every good $n$-tuple is greater than $C + \frac{\lambda}{n^{1.5}}$.