BdMO Online Forum

Define $a_i = \left | x_{i+1}-x_i \right |$ Then $\left | x_{i+j}-x_{i} \right |=\left | x_{i+j}-x_{i+j-1} \right |+\left | x_{i+j-1}-x_{i+j-2}\right |+..+\left | x_{i+1}-x_{i} \right | $
$=\sum_{k=i}^{i+j-1}a_k$
By simplifying our given inequality we can find out that
\[\left ( \sum_{i,j=1}^{n}\left | x_i-x_j \right | \right )^2\leq\frac{2\left ( n^2-1 \right )}{3} \sum_{i,j=1}^{n}(x_i-x_j)^2\]\[\Leftrightarrow \left ( \sum_{i=1}^{n-1}\sum_{j=i+1}^{n}\left | x_j-x_i \right | \right )^2\leq\frac{\left ( n^2-1 \right )}{3} \left ( \sum_{i=1}^{n-1}\sum_{j=i+1}^{n}(x_j-x_i)^2 \right )\]
\[\Leftrightarrow \left ( \sum_{i=1}^{n-1}\sum_{j=i+1}^{n}\sum_{k=i}^{j-1}a_k \right )^2\leq \left ( \frac{n^2-1}{3} \right )\left ( \sum_{i=1}^{n-1}a_i^2 +\sum_{i=1}^{n-2}(a_i+a_{i+1})^2+..(a_1+a_2+..+a_{n-1})^2 \right )\]
Sorry but now i'm jumping to:
\[\Leftrightarrow \left ( \sum_{i=1}^{n-1}a_i i(n-i) \right )^2\leq \frac{n^2-1}{3} \left ( \sum_{n-1}^{i}a_i^2i(n-i) \right )+ \frac{n^2-1}{3} \sum_{i=1}^{n-2}\sum_{j=i+1}^{n-1} 2a_i a_j i(n-j)\]
(Although i've calculate these correctly)
Again a big jump to

$\Leftrightarrow \sum_{i=1}^{n-2}\sum_{j=i+1}^{n-1}\{2a_i a_j(ij(n-i)(n-j)-\frac{n^2-1}{3}i(n-j))\}\leq $

$ \sum_{i=1}^{n-1}a_i^2( \left ( \frac{n^2-1}{3} \right )i(n-i)-i^2(n-i)^2)$

Now by A.M-G.M:

$\sum_{i=1}^{n-2}\sum_{j=i+1}^{n-1}\{2a_i a_j(ij(n-i)(n-j)-\frac{n^2-1}{3}i(n-j))\}\leq$

$ \sum_{i=1}^{n-2}\sum_{j=i+1}^{n-1}(a_i^2+a_j^2)(ij(n-i)(n-j)-\frac{n^2-1}{3}i(n-j))\}$

Now the ultimate jump of the calculations:

$\sum_{i=1}^{n-1}a_i^2( \left ( \frac{n^2-1}{3} \right )i(n-i)-i^2(n-i)^2)=\sum_{j=1}^{i-1}a_i^2\left ( ij(n-i)

(n-j)-\left ( \frac{n^2-1}{3} \right )j(n-i) \right )+$

$\sum_{j=i+1}^{n-1}a_i^2\left ( ij(n-i)(n-j)-\left ( \frac{n^2-1}{3} \right )i(n-j) \right )$

Because of A.M-G.M equality holds if and only if all $a_i$ are same which means $\{x_i\}$ is an arithmetic
progression.

I wish i could show you my calculations... But i've justified my calculations.