A finite sequence of positive integers $a_0, a_1, \dots, a_n$ is said to be logarithmically concave if $$a_i^2 \ge a_{i-1} a_{i+1}$$ for all $1 \le i \le n-1$.
Now consider the set of inequalities: $$a_ia_j \ge a_h a_k$$ for $0 \le h \le i \le j \le k \le n$ and $h + k = i + j$.
How to prove the equivalence of the two sets of inequalities given above.
Kindly share your thoughts.
Thank you.
In the forward direction, apply induction to $a_i a_j \geq a_{i - 1} a_{j + 1}$ for $0 < i \leq j < n$.
The latter, for $0 < i < j < n$, follows from $$\prod_{k = i}^{j} a^2_k \geq \prod_{k = i}^{j} a_{k - 1} a_{k + 1} = a_{i - 1} \left(a_i a_j \prod_{k = i + 1}^{j - 1}a^2_k \right) a_{j + 1}$$ (where the product of zero terms is considered to be $1$ - just divide both sides by the parenthesized expression). The other direction of the equivalence is just taking a particular case.
UPDATE. The induction application mentioned above produces $a_i a_j \geq a_{i - d} a_{j + d}$ for $0 \leq d \leq i \leq j \leq n - d$, the induction being on $d$. This is exactly what is needed: if $0 \leq h \leq i \leq j \leq k \leq n$ and $h + k = i + j$, then $h = i - d$ and $k = j + d$ with $d = i - h$.