Given the function $$f(x) = \prod_{i=1}^k x_i \prod_{j=k+1}^n (1-x_j) + \prod_{i=1}^k (1-x_i) \prod_{j=k+1}^n x_j$$ with $\frac{1}{2} < x_i < 1$ and $k > \frac{n}{2}$, I wish to produce a rearrangement inequality. It seems pretty intuitive that you'd want all the largest $x_i$ to be in places $\left\{1, \ldots, k\right\}$. In other words:

Given the numbers $a_1 \geq a_2 \geq \dots \geq a_n$, with $\frac{1}{2} < a_i < 1$ prove that $$f(a_1,a_2,\dots ,a_n) \geq f(a_{\sigma(1)},a_{\sigma(2)},\dots ,a_{\sigma(n)})$$ for any permutation $\sigma$.


Let $(x_1,\dots, x_n)$ be a permutation of $(a_1,\dots, a_n)$. Then

$$f(x) \prod_{i=1}^n \frac 1{a_i} = f(x) \prod_{i=1}^n\frac 1{x_i}= \prod_{j=k+1}^n \left(\frac 1{x_j}-1\right) + \prod_{i=1}^k\left(\frac 1{x_i}-1\right)=\prod_{j=k+1}^n y_j + \prod_{i=1}^k y_i,$$

where $y_i=\frac 1{x_i}-1$ for each $i$, so $0<y_i<1$. Put $\prod_{j=k+1}^n y_j=B$ and $\prod_{i=1}^k y_i=A$. Since $AB= \prod_{i=1}^n \left(\frac 1{a_i}-1\right)=\operatorname{const}$, sum $A+B$ is maximal when one of the summand achieves the smallest possible value. Since $k>\frac n2$ this holds when the numbers $\{y_1,\dots, y_k\}$ are the smallest $y_i$’s, that is iff the numbers $\{x_1,\dots, x_k\}$ are the largest $a_i$’s.


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