Effectively, we have to prove \[ \dfrac{n(n-1)}{2}\sum_{i<j} x_i\,x_j -\left(\sum_{i=1}^{n-1} (n-i) x_i\right)\cdot\left(\sum_{j=2}^n (j-1)\, x_j\right)>0 \] For each $ i$ satisfying $ 1\leq i\leq (n-1)$, let $ a_i = x_{i+1}+x_{i+2} + \ldots + x_n$ and let $ b=\sum_{j=2}^n (j-1)\, x_j$, and $ c_i = \dfrac{n(n-1)}{2} a_i - (n-i)b$. The LHS $ =\dfrac{n(n-1)}{2}\sum_{i<j} x_i\,x_j -\left(\sum_{i=1}^{n-1} (n-i) x_i\right)b$ $ =\dfrac{n(n-1)}{2}\sum_{i=1}^{n-1}x_ia_i -\left(\sum_{i=1}^{n-1} (n-i) x_i\right)b$ $ =\sum_{i=1}^{n-1}\left(\dfrac{n(n-1)}{2}a_i - (n-i)b\right)x_i$ $ =\sum_{i=1}^{n-1}c_ix_i$ So, it remains now to prove that $ \sum_{i=1}^{n-1}x_ic_i >0$. Since $ \sum_{i=1}^{n-1}a_i = b$ and obviously, $ \sum_{i=1}^{n-1} (n-i)=\dfrac{n(n-1)}{2}$, we obtain, $ \sum_{i=1}^{n-1}c_i = 0$. Also, since $ x_i<x_{i+1}$, we get \[ b = \sum_{j=2}^n (j-1)\, x_j < \sum_{j=2}^n (j-1)\, x_n =\dfrac{n(n-1)}{2}x_n \] So, $ c_{n-1}=\dfrac{n(n-1)}{2}x_n - b >0$. Further, we have \[ \dfrac{c_{i+1}}{n-(i+1)}-\dfrac{c_i}{n-i} = \dfrac{n(n-1)}{2}\left(\dfrac{a_{i+1}}{n-(i+1)} -\dfrac{a_i}{n-i}\right)>0 \] which implies that \[ \dfrac{c_1}{n-1}<\dfrac{c_2}{n-2}<\cdots < \dfrac{c_{n-1}}{1} \]Therefore, there exists a $ k$, such that $ c_1$, $ c_2$, $ \ldots$, $ c_k\leq 0$ and $ c_{k+1}$, $ c_{k+2}$, $ \ldots$, $ c_{n-1} > 0$. But then $ c_i(x_i - x_k)\geq 0$ for all $ i$, i.e. $ c_ix_i \geq c_ix_k$ for all $ i$. So, \[ \sum_{i=1}^{n-1} c_i x_i > \sum_{i=1}^{n-1} c_i x_k = x_k\sum_{i=1}^{n-1} c_i=0 \] as required.

A fantastic solution

Here is a solution by induction and shifting. Let P(n) be the problem statement. Base Step: $3(x_1x_2+x_1x_3+x_2x_3) > (2x_1+x_2)(x_2+2x_3) \to (x_3-x_2)(x_2-x_1) > 0$ which is true. Inductive Step: Assume that P(n) is true. We will prove that P(n+1) is true. Let $f(x_1,x_2,\dots,x_{n+1}) = \frac{n(n-1)}{2} \sum_{i < j} x_ix_j - \left(\sum^{n-1}_{i=1} (n-i)\cdot x_i \right) \cdot \left(\sum^{n}_{j=2} (j-1) \cdot x_j \right)$ It is easy to show that $f(x_1+k,x_2+k,\dots,x_{n+1}+k) = f(x_1,x_2,\dots,x_{n+1})$ So we can assume that $x_1 = 0$ and $x_2,x_3,\dots,x_{n+1} > 0$ It remains to show that $\frac{n(n+1)}{2} \sum_{2\leq i < j \leq n+1}x_ix_j > ((n-1)x_2+(n-2)x_3+\cdots+x_n)(x_2+2x_3+\cdots+nx_{n+1})$ From P(n) is true, we have $\frac{n(n-1)}{2} \sum_{2 \leq i < j \leq n+1}x_ix_j > ((n-1)x_2+(n-2)x_3+\cdots+x_n)(x_3+2x_4+\cdots+(n-1)x_{n+1}$ So, it is sufficient to show that $(n+1)(x_3+2x_4+\cdots+(n-1)x_{n+1}) > (n-1)(x_2+2x_3+\cdots+nx_{n+1})$ which is true because of $x_2 < x_3 < \cdots < x_{n+1}$