# How prove this inequality $\sum\limits_{1\le i<j\le n}ix_{i}x_{j}\le\frac{n-1}{3}\sum\limits_{k=1}^{n}kx_{k}$

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 $$\dfrac{n(n+1)(2n+1）}{6}-\dfrac{2}{n(n+1)}A^2\le\dfrac{n-1}{3}A$$ It seem not true, so my idea can't work,and How to prove this inequality? Thank you

• Indeed, the last inequality is not true, for instance, for $A=0$. – Alex Ravsky Aug 18 '13 at 2:44
• Proving for $n=2$ is trivial. Now proceed by induction on $n$. I can write out all the details if necessary. – John M Aug 18 '13 at 5:59
• Note that $\sum k x_k = \sum k x_k^2$. – marty cohen Aug 18 '13 at 6:43
• @math110 - was the below solution sufficient for you? Or do you need more detail? – John M Aug 19 '13 at 21:09

For more convenient notation, let $L_n := \sum_{1 \leq i<j\leq n} i x_i x_j$ and $R_n := \frac{n-1}{3} \sum_{1 \leq i \leq n} i x_i$. We need to prove that $L_n \leq R_n$.
We proceed by induction, the case $n = 1$ being trivial. Suppose that the claim holds for $n$, we will prove it for $n+1$. It will suffice to show that $\Delta L_n \leq \Delta R_n$ where $\Delta L_n := L_{n+1} - L_n$ and $\Delta R_n := R_{n+1} - R_n$.
We have $$\Delta L_n = \sum_{1 \leq i\leq n} i x_i x_{n+1}$$ and $$\Delta R_n = \frac{1}{3} \sum_{1 \leq i\leq n} i x_i x_{n+1} + \frac{(n+1)n}{3} x_{n+1} = \frac{1}{3} L_n + \frac{2}{3}{n+1 \choose 2} x_{n+1}$$. Thus, $\Delta R_n \geq \Delta L_n$ is equivalent to ${n+1 \choose 2} x_{n+1} \geq \Delta L_n$. But this is clear because: $$\Delta L_n = \sum_{1 \leq i\leq n} i x_i x_{n+1} \leq x_{n+1} \sum_{1 \leq i\leq n} i = x_{n+1} {n+1 \choose 2}$$