First we prove: Lemma: $ a_{1}a_{n-1}\ge a_{0}a_{n}$. Proof: Take the inequalities $ a_{i}\ge a_{i-1}a_{i+1}$ for $ i = 1,2,\cdots,n-1$ Multiplying these together and canceling factors gives $ a_{1}a_{n-1}\ge a_{0}a_{n}$. Corollary: $ a_{k}a_{n-k}\ge a_{0}a_{n}$ for $ 1\le k\le\frac{n}{2}$ We induct on $ k$. For $ k = 1$ it is the same as the lemma. If it holds for $ k$, where $ 1\ge k\le\frac{n}{2}-1$, then $ a_{k},...,a_{n-k}$ is also a log concave sequence and thus $ a_{k+1}a_{n-k-1}\ge a_{k}a_{n-k}$. By the induction hypothesis, $ a_{k}a_{n-k}\ge a_{0}a_{n}$ and thus $ a_{k+1}a_{n-k-1}\ge a_{0}a_{n}$ and the induction is complete. From this corollary, we now establish the inequality: $ (a_{1}a_{2}\cdots a_{n-1})^{2}\ge (a_{0}a_{n})^{n-1}$ We take the inequalities $ a_{k}a_{n-k}\ge a_{0}a_{n}$ for $ k = 1,...,\frac{n}{2}$. If $ n$ is odd we square each of the $ \frac{n-1}{2}$ inequalities, multiply them together, and obtain the above inequality. If $ n$ is even, then we have $ n-k = k$ for $ k =\frac{n}{2}$, so we proceed as above without squaring this particular inequality, and produce the desired result. From this we rearrange to get: $ (a_{0}a_{1}\cdots a_{n-1}a_{n})^{\frac{1}{n+1}}(a_{1}\cdots a_{n-1})^{\frac{1}{n-1}}\ge a_{0}a_{n}$ By AM-GM, we have that: \[ \frac{a_{0}+a_{1}+...+a_{n-1}+a_{n}}{n+1}\ge (a_{0}a_{1}\cdots a_{n-1}a_{n})^{\frac{1}{n+1}}\] \[ \frac{a_{1}+...+a_{n-1}}{n-1}\ge (a_{1}\cdots a_{n-1})^{\frac{1}{n-1}}\] Multiplying these together produces: \[ \frac{(a_{0}+a_{1}+...+a_{n-1}+a_{n})(a_{1}+...+a_{n-1})}{(n+1)(n-1)}\ge (a_{0}a_{1}\cdots a_{n-1}a_{n})^{\frac{1}{n+1}}(a_{1}\cdots a_{n-1})^{\frac{1}{n-1}}\ge a_{0}a_{n}\] Adding $ (a_{0}+a_{1}+...+a_{n-1}+a_{n})(a_{1}+...+a_{n-1})$ to each side, we get: \[ \frac{n^{2}(a_{0}+a_{1}+...+a_{n-1}+a_{n})(a_{1}+...+a_{n-1})}{(n+1)(n-1)}\ge a_{0}a_{n}+(a_{0}+a_{1}+...+a_{n-1}+a_{n})(a_{1}+...+a_{n-1}) = (a_{0}+\cdots+a_{n-1})(a_{1}+\cdots+a_{n})\] or \[ \frac{a_{0}+\cdots+a_{n}}{n+1}\cdot\frac{a_{1}+\cdots+a_{n-1}}{n-1}\geq\frac{a_{0}+\cdots+a_{n-1}}{n}\cdot\frac{a_{1}+\cdots+a_{n}}{n}.\] as desired.

Lemma 1: First we will prove, by induction, that $ \frac {a_{k}^{k + 1}}{a_{k + 1}^k} \ge a_0 \forall k \in \mathbb{N}$. For our base case we use $ k = 1$. Since $ a_{i - 1}a_{i + 1} \le a_i^2$ for all $ i \in \mathbb{N}$, we have $ a_0a_2 \le a_1^2 \Rightarrow \frac {a_1^2}{a_2^1} \ge a_0$. Now, suppose that $ \frac {a_k^{k + 1}}{a_{k + 1}^k} \ge a_0$ for some $ k = j$. We wish to prove it is true for $ k = j + 1$ as well. $ \dfrac{a_j^{j + 1}}{a_{j + 1}^j} \ge a_0$ $ a_{j + 1}^{j + 2} \ge \dfrac{a_{j + 1}^{2(j + 1)}}{a_j^{j + 1}} a_0$ $ a_{j + 1}^{j + 2} \ge a_{j + 2}^{j + 1}a_0$ (since $ a_ja_{j + 2} \le a_{j + 1}^2$) $ \dfrac{a_{j + 1}^{j + 2}}{a_{j + 2}^{j + 1}} \ge a_0$ Thus, the statement is true for all natural numbers $ k$. $ \Box$ Lemma 2: Now we will prove, by induction, that $ a_1a_2a_3 \ldots a_{n - 1} \ge (a_0a_n)^\frac {n - 1}{2}$ for all natural numbers $ n \ge 2$. For our base case we use $ n = 2$. We know that $ a_1^2 \ge a_0a_2$ from the problem condition, so $ a_1 \ge (a_0a_2)^\frac {1}{2}$. Now, suppose that $ a_1a_2a_3 \ldots a_{n - 1} \ge (a_0a_n)^\frac {n - 1}{2}$ for some $ n = m$. We wish to prove it is true for $ n = m + 1$ as well. $ a_1a_2a_3 \ldots a_{m - 1} \ge (a_0a_m)^\frac {m - 1}{2}$ $ a_1a_2a_3 \ldots a_m \ge (a_0a_m)^\frac {m - 1}{2} a_m$ $ a_1a_2a_3 \ldots a_m \ge a_0^\frac {m - 1}{2} a_m^\frac {m + 1}{2}$ By our first lemma, $ \dfrac{a_m^{m + 1}}{a_{m + 1}^m} \ge a_0$ $ \dfrac{a_m^\frac {m + 1}{2}}{a_{m + 1}^\frac {m}{2}} \ge a_0^\frac {1}{2}$ $ 1 \ge \dfrac{a_0^\frac {1}{2} a_{m + 1}^\frac {m}{2}}{a_m^\frac {m + 1}{2}}$ Multiplying, we get: $ a_1a_2a_3 \ldots a_m \ge a_0^\frac {m}{2} a_{m + 1}^\frac {m}{2} = (a_0a_m)^\frac {(m + 1) - 1}{2}$ This was the form we wanted; so for all natural numbers $ n \ge 2$, we have $ a_1a_2a_3 \ldots a_{n - 1} \ge (a_0a_n)^\frac {n - 1}{2}$. $ \Box$ We can write the result of our second lemma as: $ (a_1a_2 \ldots a_{n - 1})^\frac {2}{n - 1} \ge a_0a_n$ $ (a_1a_2 \ldots a_{n - 1})^\frac {2n}{n^2 - 1} \ge (a_0a_n)^\frac {n}{n + 1}$ $ (a_0a_n)^\frac {1}{n + 1} (a_1a_2 \ldots a_{n - 1})^\frac {2n}{n^2 - 1} \ge a_0a_n$ $ \sqrt [n^2 - 1]{a_0^{n - 1}a_1^{2n}a_2^{2n}a_3^{2n} \ldots a_{n - 1}^{2n} a_n^{n - 1}} \ge a_0a_n$ By the Arithmetic Mean - Geometric Mean Inequality, we get: $ \dfrac{(a_0 + a_1 + \ldots + a_n)(a_1 + a_2 + \ldots + a_{n - 1})}{n^2 - 1} \ge a_0a_n$ $ \dfrac{(a_0 + a_1 + \ldots + a_{n - 1})(a_1 + a_2 + \ldots + a_{n - 1})}{n^2(n^2 - 1)} \ge \dfrac{a_0a_n}{n^2} - \dfrac{a_n(a_1 + a_2 + \ldots + a_{n - 1})}{n^2(n^2 - 1)}$ $ \dfrac{(a_0 + a_1 + \ldots + a_{n - 1})(a_1 + a_2 + \ldots + a_{n - 1})}{n^2(n^2 - 1)} \ge \dfrac{a_0a_n}{n^2} + \dfrac{a_n(a_1 + a_2 + \ldots + a_{n - 1})}{n^2} - \dfrac{a_n(a_1 + a_2 + \ldots + a_{n - 1})}{n^2 - 1}$ $ \dfrac{(a_0 + \ldots + a_{n - 1})(a_1 + \ldots + a_{n - 1})}{n^2 - 1} + \dfrac{a_n(a_1 + \ldots + a_{n - 1})}{n^2 - 1} \ge \dfrac{(a_0 + \ldots + a_{n - 1})(a_1 + \ldots + a_{n - 1})}{n^2} + \dfrac{a_n(a_0 + \ldots + a_{n - 1})}{n^2}$ $ \left( \dfrac{a_0 + \ldots + a_{n - 1}}{n + 1} + \dfrac{a_n}{n + 1} \right) \left( \dfrac{a_1 + \ldots + a_{n - 1}}{n - 1} \right) \ge \left( \dfrac{a_0 + \ldots + a_{n - 1}}{n} \right) \left( \dfrac{a_1 + \ldots + a_{n - 1}}{n} + \dfrac{a_n}{n} \right)$ $ \dfrac{a_0 + a_1 + \ldots + a_n}{n + 1} \times \dfrac{a_1 + a_2 + \ldots + a_{n - 1}}{n - 1} \ge \dfrac{a_0 + a_1 + \ldots + a_{n - 1}}{n} \times \dfrac{a_1 + a_2 + \ldots + a_n}{n}$

This inequality, after algebraic manipulation, makes us prove that $\frac{(a_{0}+...a_{n-1})(a_{1}+...a_{n})}{n^{2}} \ge a_{0}a_{n}$. Lemma 1. We show that $a_{0}a_{n} \le a_{i}a_{n-i}$. We have the inequalities: $a_{i}a_{i+2} \le a_{i+1}^{2}$, $a_{i+1}a_{i+3} \le a_{i+2}^{2}$, ... $a_{n-2}a_{n} \le a_{n-1}^{2}$. Multiplying all the inequalities and dividing, we get the desired result. Now, from our original LHS, we see that $\frac{(a_{0}+...a_{n-1})(a_{1}+...a_{n})}{n^{2}} \ge ((a_{0}...a_{n-1})(a_{1}...a_{n}))^{1/n}$, which is $((a_{0}a_{n})(a_{1}a_{n-1})...(a_{n-1}a_{1}))^{1/n} \ge a_{0}a_{n}$.