# Given reals $a_1, a_2, \cdots, a_{n - 1}, a_n$ such that $\sum_{i = 1}^na_1^2 = 1$. Calculate the maximum value of $\sum_{cyc}|a_1 - a_2|$.

Given reals $$a_1, a_2, \cdots, a_{n - 1}, a_n$$ such that $$a_1^2 + a_2^2 + \cdots + a_{n - 1}^2 + a_n^2 = 1$$ $$(n \in \mathbb N, n \ge 3)$$. Calculate the maximum value of $$\large |a_1 - a_2| + |a_2 - a_3| + \cdots + |a_{n - 1} - a_n| + |a_n - a_1|$$

There must exist $$1 < k < n$$ such that $$a_{k - 1} \le a_k \le a_{k + 1}$$

$$\implies |a_{k + 1} - a_k| + |a_k - a_{k + 1}| = |a_{k + 1} - a_{k + 1}|$$.

Repeat the above process for about $$n - 1$$ times and we have that $$|a_1 - a_2| + |a_2 - a_3| + \cdots + |a_{n - 1} - a_n| + |a_n - a_1| \le 2 \cdot \min(|a_i - a_j|, 1 \le i < j \le n)$$

Now we just have to find the maximum value of $$\min(|a_i - a_j|, 1 \le i < j \le n)$$ for $$a_1^2 + a_2^2 + \cdots + a_{n - 1}^2 + a_n^2 = 1$$, which I don't know how.

The maximal value is $$2\sqrt{n-1}$$ if $$n$$ is odd, and $$2\sqrt{n}$$ if $$n$$ is even. We can prove the following:

Let $$a_1, \ldots, a_n$$ be real numbers, $$n \ge 2$$. Then $$\tag{*} |a_1 - a_2| + |a_2 - a_3| + \ldots + |a_{n - 1} - a_n| + |a_n - a_1| \le c_n \sqrt{a_1^2 + \ldots + a_n^2}$$ where $$c_n = 2\sqrt{n-1}$$ if $$n$$ is odd, and $$c_n = 2\sqrt{n}$$ if $$n$$ is even. The bounds are sharp.

Proof: Case 1: $$n$$ is even. Then $$|a_1 - a_2| + |a_2 - a_3| + \ldots + |a_{n - 1} - a_n| + |a_n - a_1| \\ \underset{(1)}{\le} \sum_{k=1}^n (|a_k| + |a_{k+1}|) = 2 \sum_{k=1}^n (1 \cdot |a_k|) \underset{(2)}{\le} 2 \sqrt{n} \sqrt{\sum_{i=1}^{n} a_i^2 } \, ,$$ where the last step uses the Cauchy-Schwarz inequality.

Equality holds at $$(1)$$ if the $$a_k$$ have alternating signs, and equality holds at $$(2)$$ if all $$|a_k|$$ are equal. It follows that equality holds in $$(*)$$ exactly if $$(a_1, \ldots, a_n) = (x, -x, \ldots, x, -x)$$ for some $$x \in \Bbb R$$.

Case 2: $$n$$ is odd. There must be (at least) one index $$k$$ such that $$a_{k-1} - a_k$$ and $$a_k - a_{k+1}$$ have the same sign. Without loss of generality $$k=n$$, so that $$|a_{n-1} - a_n | + |a_n - a_{1}| = |a_{n-1} - a_{1}| \, .$$ Then, using the already proven estimate for the even number $$n-1$$, $$|a_1 - a_2| + |a_2 - a_3| + \ldots + |a_{n - 1} - a_n| + |a_n - a_1| \\ = |a_1 - a_2| + |a_2 - a_3| + \ldots + |a_{n - 1} - a_1| \\ \underset{(3)}{\le} 2\sqrt{n-1} \sqrt{\sum_{i=1}^{n-1} a_i^2 } \underset{(4)}{\le} 2\sqrt{n-1} \sqrt{\sum_{i=1}^{n} a_i^2 } \, .$$

Equality holds at $$(3)$$ if $$(a_1, \ldots, a_{n-1}) = (x, -x, \ldots, x, -x)$$, and equality at $$(4)$$ holds if $$a_n = 0$$. It follows that equality holds in $$(*)$$ exactly if $$(a_1, \ldots, a_n) = (x, -x, \ldots, x, -x, 0)$$ for some $$x \in \Bbb R$$, or a cyclic rotation thereof.

• Big mistake on my solution there. Thanks for (indirectly) pointing that out. – Lê Thành Đạt Jan 20 at 10:28
• @LêThànhĐạt: I have simplified the proof further. After doing that it looks quite similar to what you wrote (+1 to your answer). I hope that you'll believe that it was developed independently. – Martin R Jan 20 at 10:49
• I really do believe so. (What I had written in the description under the question was just a false idea I made up. The logic falls apart really quickly if you read my "attempt" more carefully.) – Lê Thành Đạt Jan 20 at 10:51

We have that $$|x - y| = 2 \cdot \max(x, y) - (x + y)$$

$$\implies \sum_{cyc}|a_1 - a_2| = 2 \cdot \left[\sum_{cyc}\max(a_1, a_2) - \sum_{i = 1}^na_1\right]$$

which could be rewritten as $$\sum_{cyc}|a_1 - a_2| = 2 \cdot \sum_{i = 1}^nx_ia_i$$ where $$x_i \in \{-1, 0, 1\}, i = \overline{1, n}$$ and $$\displaystyle\sum_{i = 1}^nx_i = 0$$.

In the case of $$n$$ being odd-numbered, there must exist $$m$$ $$(1 \le m \le n)$$ such that $$x_m = 0$$, otherwise $$\displaystyle\sum_{i = 1}^nx_i$$ would be odd.

Let $$x_n = 0$$, we obtain that $$\sum_{cyc}|a_1 - a_2| \le 2 \cdot \sum_{i = 1}^{n - 1}x_ia_i \le 2 \cdot \sum_{i = 1}^{n - 1}|a_i| \le 2\sqrt{(n - 1) \cdot \sum_{i = 1}^{n - 1}a_i^2} = 2\sqrt{n - 1}$$

The equality sign occurs when $$a_i = \pm \sqrt{\dfrac{1}{n - 1}}, i = \overline{1, n - 1}$$ and $$a_n = 0$$ such that $$\displaystyle\sum_{i = 1}^{n}a_1 = 0$$.

The same progress could be done for even-numbered values of $$n$$.