Simple algebra involving bounding of a sequence

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Simple algebra involving bounding of a sequence

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Source: 2025 IRN-MNG Friendly Competition

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Iveela

#1 h Jun 8, 2025, 10:29 AM • 1 Y

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Let $n \geq 2$ be a positive integer. Let $x_1, \dots, x_n$ be a sequence of real numbers such that
$\max_{1 \leq k \leq n} \frac{x_1 + \dots + x_k}{k} = 1 \quad \text{and} \quad \min_{1 \leq k \leq n} \frac{x_1 + \dots + x_k}{k} = 0.$ Find the minimum and maximum possible values of $\max_{1 \leq k \leq n}x_k - \min_{1 \leq k \leq n}x_k$ .

This post has been edited 2 times. Last edited by Iveela, Jun 8, 2025, 12:53 PM

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zhaoli

#2 h Jun 8, 2025, 12:31 PM

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Need to clear:
$\max_{1 \leq k \leq n} \frac{x_1 + \dots + x_k}{k}=?$

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#3 h Jun 10, 2025, 3:38 PM

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Part 1: The minimum possible value of $\max_{1\leq i\leq n}x_k-\min_{1\leq k\leq n}x_k$ is $\frac{n}{n-1}$ .

Solution. We first show that this value is attainable. By setting $x_1=0$ and $x_i=\frac{n}{n-1}$ for $2\leq i\leq n$ , both conditions are met with the desired range. We now show that this is the smallest possible attainable value. By the conditions there must exist indices $k_1$ and $k_0$ such that $x_1+\dots+x_{k_1}=k_1\quad\text{and}\quad x_1+\dots+x_{k_0}=0.$
We first suppose that $k_1<k_0$ . One of $x_1,\dots,x_{k_1}$ must be at least $1$ so the the maximum of the sequence is at least $1$ . On the other hand, $x_{k_1+1}+\dots+x_{k_0}=-k_1.$ Hence one of these terms must be less than $\frac{-k_1}{k_0-k_1}\leq\frac{-1}{k_0-1}\leq \frac{-1}{n-1}$ . Thus the maximum minus the minimum in this case is at least $\frac{n}{n-1}$ as wanted.

Now suppose $k_0<k_1$ . Then we have $x_{k_0+1}+\dots+x_{k_1}=k_1.$ So one of these terms is at least $\frac{k_1}{k_1-k_0}\geq \frac{n}{n-1}$ . Also since the sum of the first $k_0$ terms is $0$ , one of them must be at most $0$ , finishing this part of the problem.

Part 2: The maximum possible value of $\max_{1\leq i\leq n}x_k-\min_{1\leq k\leq n}x_k$ is $2n-2$ .

Solution. We first show that this value is attainable. By setting $x_{n-1}=n-1$ , $x_n=-(n-1)$ , and $x_i=0$ for $1\leq i\leq n-2$ , both conditions are met with the desired range. We now show that this is the largest possible attainable value. Begin by noticing that $-(k-1)\leq x_k\leq k$ for all $k$ . Suppose $x_{i_1}=M$ is the maximum and $x_{i_2}=m$ is the minimum. Then $M-m\leq i_1+(i_2-1)\leq 2n-2 ,$ finishing the problem.

This post has been edited 1 time. Last edited by sami1618, Jun 10, 2025, 3:38 PM

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