$\frac{a_1 + 1}{2} \cdot \frac{a_2 + 2}{2} \cdot \frac{a_3 + 3}{2} \cdot \frac{a_4 + 4}{2} \cdot \frac{a_5 + 5}{2} \cdot \frac{a_6 + 6}{2} > 6!.$

Question

Find the number of permutations $(a_1, a_2, a_3, a_4, a_5, a_6)$ of $(1,2,3,4,5,6)$ that satisfy $$\frac{a_1 + 1}{2} \cdot \frac{a_2 + 2}{2} \cdot \frac{a_3 + 3}{2} \cdot \frac{a_4 + 4}{2} \cdot \frac{a_5 + 5}{2} \cdot \frac{a_6 + 6}{2} > 6!.$$

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I tried to find some permutations that work and then find a pattern, but that didn't really work. I also tried using complementary counting, finding the ones that don't work, and then subtracting them from the total number of permutations, but that also failed. I'm not sure what else I can do, and any guidance would be greatly appreciated!!

Thanks in advance!!!

Answer 1

The product on the left is equal to $6!$ if $(a_1, \ldots, a_6) = (1, \ldots, 6)$.

Here are two completely different ways to show that the product is strictly larger than $6!$ for any other permutation:

Approach #1: The inequality between arithmetic and geometric mean shows that $$ \frac{a_j+j}{2} \ge \sqrt{j a_j} \, $$ with equality if and only if $a_j = j$. This shows that the product on the left is $$ \ge \sqrt{(1 a_1) (2 a_2) \cdots (6 a_6)} = 6! \, $$ with equality if and only if $a_j = j$ for $1 \le j \le 6$.

Approach #2: Assume that $a_j > a_k$ for some $j < k$. Then $$ (a_j+j)(a_k+k) - (a_k + j)(a_j + k) = (a_j-a_k)(k-j) > 0 \\ \implies (a_j+j)(a_k+k) > (a_k + j)(a_j + k) \, , $$ so that the product on the left becomes smaller if $a_j$ and $a_k$ are exchanged.

Therefore, if the $a_j$ are not in increasing order, the product decreases if the smallest $a_j$ is moved to the first position, then the second smallest $a_j$ is moved to the second position, and so on.

This also shows that product is strictly larger than $6!$ if the $a_j$ are not in increasing order.