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

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

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!!

• The minimum of the LHS is achieved at the identity permutation.
– lhf
Commented Feb 25, 2022 at 17:01
• So only $(1,2,3,4,5,6)$ doesn't work? Commented Feb 25, 2022 at 17:02
• So it seems.... And the minimum is exactly $6!$.
– lhf
Commented Feb 25, 2022 at 17:19
• There is a famous inequality which you can apply to every factor. Commented Feb 25, 2022 at 17:33

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.