Inequality involving an increasing sequence

Question

Let $(x_1,x_2,\dots,x_n)$ be an increasing sequence of numbers in $[0,1]$. I'd like to show that there exists a universal constant $ c > 0 $ such that $$ \sum_{i=1}^{n-1}\sqrt{i(n-i)}(x_{i+1}-x_i) \leq c\sqrt{\sum_{i,i'}(x_i - x_{i'})^2}. $$ It seems plausible since we have the identity $$ \tag{1} \sum_{i=1}^{n-1}i(n-i)(x_{i+1}-x_i) = \frac{1}{2}\sum_{i,i'}|x_i - x_{i'}|. $$ Edit: Proof of (1). Using the fact the sequence of numbers is increasing, we have $$ \sum_{i,i'}|x_i - x_{i'}| = 2\sum_{i'\leq i}(x_i - x_{i'}). $$ Next, $ x_i - x_{i'} = \sum_{j = i'}^{i-1}(x_{j+1}-x_j) $. Hence, \begin{align*} \sum_{i,i'}|x_i - x_{i'}| & = 2\sum_{i'\leq i}\sum_{j = i'}^{i-1}(x_{j+1}-x_j) \\ & = 2\sum_{i'=1}^n \sum_{i=i'}^n\sum_{j = i'}^{i-1}(x_{j+1}-x_j) \\ & = 2\sum_{i'=1}^n \sum_{j=i'}^{n-1}\sum_{i = j+1}^{n}(x_{j+1}-x_j) \\ & = 2\sum_{j=1}^{n-1} \sum_{i'=1}^{j}\sum_{i = j+1}^{n}(x_{j+1}-x_j) \\ & = 2\sum_{j=1}^{n-1}j(n-j)(x_{j+1}-x_j). \end{align*}

Answer 1

Here's a start.

The right side can be simplified.

Let $s_1 =\sum_{j=1}^nx_{j} $ and $s_2 =\sum_{j=1}^nx_{j}^2 $,

$\begin{array}\\ r(n) &=\sum_{i=1}^n\left(x_i - \frac{1}{n}\sum_{j=1}^nx_{j}\right)^2\\ &=\sum_{i=1}^n\left(x_i - \frac{1}{n}s_1\right)^2\\ &=\sum_{i=1}^n\left(x_i^2 - \frac{2x_i}{n}s_1+\frac1{n^2}s_1^2\right)\\ &=s_2 - \frac{2}{n}s_1^2+\frac1{n}s_1^2\\ &=s_2 - \frac{1}{n}s_1^2\\ \end{array} $