Is the following property true of any repeating cycle of $n$ real values?

Question

I have found it challenging to state this inductive argument about a cycle of reals clearly and concisely.

I would greatly appreciate if someone could show me how to make the same argument in a more standard way or offer suggestions for tightening up the argument.

Let:

• $c_1, c_2, \dots, c_n$ form a repeating cycle of $n$ reals with: $$c_{i+n} = c_i$$
• $M(c_1, c_2, \dots, c_n) = \sum\limits_{i=1}^n c_i$

Claim:

There exists $s$ such that for all $j \ge 0$: $$\sum\limits_{t=0}^j(c_{s+t}) \le \dfrac{(j+1)M(c_1,\dots,c_n)}{n}$$

Argument:

(1) Since it is a cycle, if the property holds for $0 \le j \le n-1$, it follows that for all $j \ge n$

$$\sum\limits_{t=0}^j(c_{i+t}) \le \dfrac{(j+1)M(c_1,\dots,c_n)}{n}$$

(2)  Base Case:  $n=2$

• This follows since $s=\text{min}(c_1,c_2) \le \dfrac{c_1+c_2}{2}$

(3)  Assume that it is true for any such cycle consisting of up to $n \ge 2$ reals.

(4)  Inductive Case: Let $c_1, c_2, \dots, c_{n+1}$ form a cycle of $n+1$ reals with: $$c_{i+n+1} = c_i$$

(5)  There exists $k$ where $c_k \le \dfrac{M(c_1,\dots,c_{n+1})}{n+1}$

(6)  Define $d_1, d_2, \dots, d_n$ such that:

$$d_i = \begin{cases} c_{k+1} - \frac{M(c_1,\dots,c_{n+1})}{n+1} + c_{k}, & i = 1\\ c_{k+i}, & i > 1\\ \end{cases}$$

so that:

$$\frac{M(d_1, \dots, d_n)}{n} = \frac{M(c_1, \dots, c_{n+1})}{n+1}$$

(7) By assumption, there exists $i$ such that for all $j \ge 0$:

$$\sum\limits_{t=0}^j(d_{i+t}) \le \frac{(j+1)M(d_1, \dots, d_n)}{n}$$

(8) Case 1: $i = 1$

• $s=k$ for the desired property

• For $j = 0$:

$$c_k \le \frac{M(c_1,\dots,c_{n+1})}{n+1}$$

• And, for $j \ge 1$

$$\sum\limits_{t=0}^j (c_{k+t}) = \frac{M(c_1,\dots,c_{n+1})}{n+1} + \sum\limits_{t=0}^{j-1}(d_{i+t}) \le \frac{M(c_1,\dots,c_{n+1})}{n+1} + \frac{(j)M(d_1, \dots, d_n)}{n} = \frac{(j+1)M(c_1, \dots, c_{n+1})}{n+1}$$

(9)  Case 2:  $i > 1$

• $s=k+i$ for the desired property

• For $0 \le j < n-i+1$

$$\sum\limits_{t=0}^j (c_{k+i+t}) = \sum\limits_{t=0}^j (d_{i+t}) \le \frac{(j+1)M(d_1, \dots, d_n)}{n} = \frac{(j+1)M(c_1, \dots, c_{n+1})}{n+1}$$

• For $j = n-i+1$

$$\sum\limits_{t=0}^j (c_{k+i+t}) = \frac{M(c_1,\dots,c_{n+1})}{n+1} + \sum\limits_{t=0}^{j-1} (d_{i+t}) \le \frac{M(c_1,\dots,c_{n+1})}{n+1} + \frac{(j)M(d_1, \dots, d_n)}{n} = \frac{(j+1)M(c_1, \dots, c_{n+1})}{n+1}$$

• For $j \ge n-i+2$

$$\sum\limits_{t=0}^j(c_{k+i+t}) = \frac{M(c_1,\dots,c_{n+1})}{n+1} + \sum\limits_{t=0}^{j-1}(d_{i+t}) \le \frac{M(c_1,\dots,c_{n+1})}{n+1} + \frac{(j)M(d_1, \dots, d_n)}{n} = \frac{(j+1)M(c_1, \dots, c_{n+1})}{n+1}$$

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Edit:

Apologies for the rewrite in case you saw the earlier version of this argument.

I found a mistake which I believe that I have now corrected.

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Edit 2:

Made changes suggested by John Omielan.

Answer 1

Your proof is basically correct. The several small errors and a few things which I believe could be improved have now been incorporated into the question text.

You also ask about any other way to make the same argument. Here's a different, I believe simpler, way to do the proof. First, note dividing both sides of your claim by $j + 1$ gives

$$\frac{\sum\limits_{t=0}^{j}c_{s+t}}{j + 1} \le \frac{M(c_1,\dots,c_n)}{n} \tag{1}\label{eq1A}$$

This is stating there's a starting index where all of the partial sum averages are less than or equal to the cycle average. This can be shown fairly directly using the same basic technique used in this answer to Numbers $+1$, $-1$ on a circle.. First, to make it easier to explain the proof, normalize the values by subtracting $\frac{M(c_1,\dots,c_n)}{n}$ from each of the $c_i$ values. This will also reduce each partial sum average by this same amount and, in particular, it means the cycle average becomes $0$. Next, define

$$g_u = c_u - \frac{M(c_1,\dots,c_n)}{n}, \; u \ge 1 \tag{2}\label{eq2A}$$

i.e., $g_u$ is a repeating cycle of $n$ real values. Since the cycle average of $g_u$ is $0$, the partial sum averages need to be non-positive, which is equivalent to the partial sums themselves always being non-positive. Thus, \eqref{eq1A} becomes the equivalent requirement of

$$h(s, j) = \sum\limits_{t=0}^{j}g_{s+t} \le 0 \tag{3}\label{eq3A}$$

Choose any initial $1 \le s \le n$. If \eqref{eq3A} is always true, then we're done. Otherwise, as suggested by the linked answer, let $q$ be where $h(s, q)$ is maximum (if the maximum is not unique, then just choose any one). Using $s + q + 1$ instead of $s$ will now work. This is because, with the original $s$, if any partial sum starting from that point becomes positive, say with some $r \ge 0$ where $h(s + q + 1, r) \gt 0$, then $h(s, q + 1 + r) = h(s, q) + h(s + q + 1, r) \gt h(s, q)$, which contradicts $h(s, q)$ being the maximum value among the partial sums.