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Let $x_{i}$, for $i=1,2,\cdots,n$, be real numbers such that for $n\geq 2$ $$x_{1}\le x_{2}\le\cdots\le x_{n}.$$

Prove that $$\dfrac{n(n-1)}{2}\sum_{1\le i<j\le n}x_{i}x_{j}\ge\left(\sum_{i=1}^{n-1}(n-i)x_{i}\right)\left(\sum_{i=1}^{n-1}ix_{i+1}\right).$$

My idea:

Note that $$\sum_{1\le i<j\le n}x_{i}x_{j}=\sum_{i=1}^{n-1}x_{i}(x_{i+1}+x_{i+2}+\cdots+x_{n})$$ and let $$y_{i}=x_{i+1}+x_{i+2}+\cdots+x_{n}.$$

Then $$y_{1}+y_{2}+\cdots+y_{n-1}=\sum_{i=1}^{n=1}ix_{i+1}$$

following I can't have work.

This problem is from my friends.

Thank you

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up vote 3 down vote accepted

Use mathematical induction method. The inequality hold for $n=2$. Suppose it holds for $n$, then for $n+1$ we have $$ \begin{aligned} LHS_{n+1}&=\frac{n(n+1)}{2}\sum_{1\leq i<j\leq n+1}x_ix_j=\frac{n(n+1)}{2}\left(\sum_{1\leq i<j\leq n}x_ix_j+x_{n+1}\sum_{i=1}^nx_i\right)\\ &=(1+\frac{2}{n-1})LHS_n+\frac{n(n+1)}{2}x_{n+1}\sum_{i=1}^nx_i \end{aligned} $$ and $$ \begin{aligned} RHS_{n+1}&=\left(\sum_{i=1}^{n-1}(n-i)x_i+\sum_{i=1}^nx_i\right)\left(\sum_{i=1}^{n-1}jx_{j+1}+nx_{n+1}\right)\\ &=RHS_n+nx_{n+1}\sum_{i=1}^n(n+1-i)x_i+\sum_{i=1}^nx_i\sum_{j=1}^{n-1}jx_{j+1}. \end{aligned} $$ Then it follows that $$ \begin{aligned} LHS_{n+1}-RHS_{n+1}\geq& nx_{n+1}\left(\frac{n+1}{2}\sum_{i=1}^nx_i-\sum_{i=1}^n(n+1-i)x_i\right)\\&+\frac{2}{n-1}RHS_n-\sum_{i=1}^nx_i\sum_{j=1}^{n-1}jx_{j+1}\\ \text{(substitute $RHS_n$)}=&nx_{n+1}\left(\frac{n-1}{2}\sum_{i=1}^nx_i-\sum_{i=1}^{n-1}(n-i)x_i\right)\\ &-\sum_{j=1}^{n-1}jx_{j+1}\left(\sum_{i=1}^nx_i-\frac{2}{n-1}\sum_{i=1}^{n-1}(n-i)x_i\right)\\ =&\left(nx_{n+1}-\frac{2}{n-1}\sum_{j=1}^{n-1}jx_{j+1}\right)\left(\frac{n-1}{2}\sum_{i=1}^nx_i-\sum_{i=1}^{n-1}(n-i)x_i\right)\\ \stackrel{def}=& A_nB_n \end{aligned} $$ $A_n\geq0$ is due to $x_{n+1}\geq x_j$, $j=1,\dots,n$ and $B_n\geq 0$ can be derived from mathematical induction again (which is simple). This implies $LHS_{n+1}\geq RHS_{n+1}$.

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Thank you, It's nice, I think this problem have other nice methods. Thank you – math110 Aug 16 '13 at 11:55

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