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Given $x_1, x_2, x_3,..., x_n$ is a positive real number that satisfies $x_1+x_2+x_3+...+x_n=1$ for natural number $n\ge2$. Prove that $$\frac{n}{8}\ge \sum_{{1}\le{i}<{j}\le{n}} {x_i}{x_j}\ge \sum_{i=1}^n 2{({x_i}^2-x_i)^2}$$ Also find all solutions $(x_1, x_2, x_3,..., x_n, n)$ that satisfy $$\frac{n}{8}= \sum_{{1}\le{i}<{j}\le{n}} {x_i}{x_j}= \sum_{i=1}^n 2{({x_i}^2-x_i)^2}$$

My work:

I tried using Cauchy-Schwarz to get $$\left(x_1^2+x_2^2+\cdots+x_n^2\right)\left(x_2^2+x_3^2+\cdots+x_n^2+x_1^2\right)\geq(x_1 x_2 + x_2 x_3 + \dots + x_{n - 1} x_n + x_n x_1)^2,$$ but I don't think this helps, and I'm not sure how to proceed.

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2 Answers 2

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We remark that $$\left(\sum_{i=1}^nx_i\right)^2 =\sum_{i=1}^nx_i^2+ 2\sum_{{1}\le{i}<{j}\le{n}} {x_i}{x_j}$$ then $$\sum_{{1}\le{i}<{j}\le{n}} {x_i}{x_j}= \frac{1}{2}-\frac{1}{2}\sum_{i=1}^nx_i^2$$


For the left inequality, we have $$\sum_{{1}\le{i}<{j}\le{n}} {x_i}{x_j} =\frac{1}{2}-\frac{1}{2}\sum_{i=1}^nx_i^2\le \frac{1}{2}- \frac{(\sum_{i=1}^nx_i)^2}{2n}=\frac{2n-1}{2n}$$ and $$\frac{2n-1}{2n} \le\frac{n}{2} \Longleftrightarrow (n-2)^2 \qquad \text{which is true}$$ The equality occurs if and only if $n = 2$ and $x_1=x_2 = \frac{1}{2}$


For the right inequality, it is equivalent to $$\begin{align} &\Longleftrightarrow \frac{1}{2} \le \frac{1}{2}\sum_{i=1}^nx_i^2+\sum_{i=1}^n(x_i^2-x_i)^2 \\ &\Longleftrightarrow \sum_{i=1}^n \left(4x_i^4 -8x_i^3+5x_i^2 \right) \le 1 \tag{1}\\ \end{align}$$

As for all $i\in \{1,...,n\}$, we have $(2x_i-1)^2x_i(x_i-1) \le 0$, then $$4x_i^4 -8x_i^3+5x_i^2 \le x_i$$ Make the sum, we deduce that $(1)$ holds true as $$\sum_{i=1}^n \left(4x_i^4 -8x_i^3+5x_i^2 \right) \le \sum_{i=1}^n x_i =1$$ The equality occurs if and only if $(2x_i-1)^2x_i(x_i-1) = 0$ for all $i\in \{1,...,n\}$ or $x_i \in \left\{0,\frac{1}{2},1 \right\}$. Given the fact that $\sum_{i=1}^n x_i = 1$ and $x_i \ge 0$, then the equality occurs if and only if one among 2 conditions below occurs

  1. One of $x_i$ is equal to $1$ (for example, $x_1 = 1$), the other $(n-1)$ variables are equal to $0$ (for example, $x_2 =...=x_n = 0$)
  2. Or two of $x_i$ are equal to $\frac{1}{2}$ (for example, $x_1 = x_2 = \frac{1}{2}$), the other $(n-2)$ variables are equal to $0$ (for example, $x_3 =...=x_n = 0$)
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The left inequality. $$\sum_{1\leq i<j\leq n}(x_i-x_j)^2\geq0$$ gives $$(n-1)\sum_{i=1}^nx_i^2-2\sum_{1\leq i<j\leq n}x_ix_j\geq0$$ or $$(n-1)\left(\sum_{i=1}^nx_i^2+2\sum_{1\leq i<j\leq n}x_ix_j\right)\geq2n\sum_{1\leq i<j\leq n}x_ix_j$$ and from here $$\sum_{1\leq i<j\leq n}x_ix_j\leq\frac{n-1}{2n}\leq\frac{n}{8}$$ because the last inequality it's $(n-2)^2\geq0.$

The right inequality.

We need to prove that: $$\frac{1-\sum\limits_{i=1}^nx_i^2}{2}\geq\sum_{i=1}^n(2x_i^4-4x_i^3+2x_i^2)$$ or $$\sum_{i=1}^n(4x_i^4-8x_i^3+5x_i^2)\leq1$$ or $$\sum_{i=1}^n(4x_i^4-8x_i^3+5x_i^2-x_i)\leq0$$ or $$\sum_{i=1}^nx_i(x_i-1)(2x_i-1)^2\leq0$$ and we are done.

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