How prove this inequality $\sum_{1\le i

How prove this inequality $\sum_{1\le i

let $n\ge 2,n\in Z$,and $x_{1},x_{2},\cdots,x_{n}\in[0,1]$, show that
$$\sum_{1\le i<j\le n}ix_{i}x_{j}\le\dfrac{n-1}{3}\sum_{k=1}^{n}kx_{k}$$
This problem is (2013,8.16) chia west compition
my idea: let \begin{align*}
&\dfrac{n(2n+1)(n+1)}{6}=n^2+(n-1)^2+\cdots+1\\
&\ge(x_{1}+x_{2}+\cdots+x_{n})^2+(x_{2}+x_{3}+\cdots+x_{n})^2+(x_{3}+\cdots+x_{n})^2+\cdots+(x_{n-1}+x_{n})^2+x^2_{n}\\
&=nx^2_{n}+(n-1)x^2_{n-1}+\cdots+2x^2_{2}+x^2_{1}+2\sum_{1\le i<j\le
n}ix_{i}x_{j}\\ &=\sum_{k=1}^{n}kx^2_{k}+2\sum_{1\le i<j\le n}ix_{i}x_{j}
\end{align*} then
$$\sum_{k=1}^{n}kx^2_{k}\le\dfrac{n(n+1)(2n+1)}{6}-2\sum_{1\le i<j\le
n}ix_{i}x_{j}$$ use Cauchy-Schwarz inequality
$$\sum_{k=1}^{n}kx^2_{k}\sum_{k=1}^{n}k\ge(\sum_{k=1}^{n}(kx_{k})^2\Longrightarrow\sum_{k=1}^{n}kx^2_{k}\ge\dfrac{2}{n(n+1)}\left(\sum_{k=1}^{n}kx_{k}\right)^2$$
then we have
$$\dfrac{2}{n(n+1)}\left(\sum_{k=1}^{n}kx_{k}\right)^2\le\dfrac{n(n+1)(2n+1)}{6}-2\sum_{1\le
i<j\le n}ix_{i}x_{j}$$
it suffices to prove that
$$\dfrac{n(n+1)(2n+1)}{6}-\dfrac{2}{n(n+1)}\left(\sum_{k=1}^{n}kx_{k}\right)^2\le\dfrac{n-1}{3}\sum_{k=1}^{n}kx_{k}$$
then I let $$\sum_{k=1}^{n}kx_{k}=A\in[0,\dfrac{n(n+1)}{2}]$$ then it will
prove that
$$\dfrac{n(n+1)(2n+1j}{6}-\dfrac{2}{n(n+1)}A^2\le\dfrac{n-1}{3}A$$ It
seem not true, so my idea can't work,and How to prove this inequality?
Thank you