Looking for a slick way to prove this. 
Suppose that $a_k \ge 0$, $k=0,1,2,3,…$ satisfies $a_k^2\le a_{k-1}a_{k+1}$ for all $k\ge 1$.  Show that for all $k\in \{0,1,\dots, N\}$ we have  $$ a_k \le a_0^{1-k/N} a_N^{k/N} $$

I am looking for a slick way to prove this statement. I have the solution, but it is pretty messy. I need it for the solutions manual I'm helping fill in for the new edition of my professor's book (he actually asked me to post it here for a slicker solution). Also, I'm not quite sure what to tag this as? I just put number theory, but please feel free to change if you can think of a better tag. Thank you!
 A: The case $k=0$ is obvious. 
For $k\ge 1$ define $\displaystyle r_k:=\frac{a_{k}}{a_{k-1}},\quad k\ge 1$ and hence we can define $$q_k:=\ln r_k\in \mathbb{R}\quad k\ge1$$ Then your condition says that $\{q_k\}$ forms an increasing sequence of real numbers. Now define $$ s_n:=\dfrac{\sum_{k=1}^n q_k}{n}\quad n\ge 1$$ Then note that $s_n\le s_{n+1}\ \forall n\ge 1$ because $$s_{n}-s_{n+1}=\frac{(n+1)(q_1+q_2+\cdots +q_n)-n(q_1+q_2+\cdots +q_n+q_{n+1})}{n(n+1)}\\ =\frac{\frac{q_1+q_2+\cdots +q_n}{n}-q_{n+1}}{n+1}\le 0$$ The last inequality follows because $q_k\le q_{n+1}\quad \forall 1\le k\le n$. Hence, $\forall k\in \{1,2,\cdots \ ,N\}$ $$s_k\le s_{N}\\ \Rightarrow \frac{q_1+q_2+\cdots +q_k}{k}\le \frac{q_1+q_2+\cdots +q_N}{N} \\ \Rightarrow \ln \left(\displaystyle \prod_{j=1}^k r_j^{1/k}\right)\le \ln \left(\displaystyle \prod_{j=1}^N r_j^{1/N}\right)\\ \Rightarrow \left(\frac{a_k}{a_0}\right)^{1/k}\le \left(\frac{a_N}{a_0}\right)^{1/N}\Rightarrow a_k\le a_0^{1-k/N}a_N^{k/N}$$
A: Since 
$$a_k^2\le a_{k-1}a_{k+1} \quad\iff\quad \frac{a_k}{a_{k-1}} \le \frac{a_{k+1}}{a_k}$$
the sequence $a_k/a_{k-1}$ is an increasing sequence. As a result,
$$\begin{align}
\frac{a_N}{a_k} = & \frac{a_N}{a_{N-1}}\frac{a_{N-1}}{a_{N-2}}\cdots\frac{a_{k+1}}{a_k}
\ge \left(\frac{a_{k}}{a_{k-1}}\right)^{N-k}\\
\frac{a_k}{a_0} = & \frac{a_k}{a_{k-1}}\frac{a_{k-1}}{a_{k-2}}\cdots\frac{a_1}{a_0} 
\le \left(\frac{a_{k}}{a_{k-1}}\right)^k
\end{align}$$
This implies
$$\left(\frac{a_N}{a_k}\right)^k \ge \left(\frac{a_k}{a_{k-1}}\right)^{k(N-k)}
\ge \left( \frac{a_k}{a_0} \right)^{N-k}$$
which is equivalent to 
$$a_k^N \le a_N^k a_0^{N-k} \quad\iff\quad a_k  \le a_0^{1-\frac{k}{N}} a_N^{\frac{k}{N}}$$
Update
Thinking about it, this proof do assume all $a_k > 0$. If $a_k = 0$ for some $0 < k < N$, then $a_{k-1}^2 \le a_{k-2} a_{k}$ and $a_{k+1}^2 \le a_{k} a_{k+2}$ implies $a_{k+1} = a_{k-1} = 0$ whenever $k - 2$ or $k + 2$ falls within the range $[0, N]$. This implies $a_k = 0$ for every $0 < k < N$. What we need to show becomes trivial.
A: I will first assume all $a_k > 0$. First note that
$$
{a_k}^2 \leq a_{k - 1}a_{k + 1} \iff \frac{a_k}{a_{k -1}} \leq \frac{a_{k + 1}}{a_k} \\
a_k \leq {a_0}^{1 - \frac{k}{N}}{a_N}^{\frac{k}{N}} \iff {a_k}^N \leq {a_0}^{N - k}{a_N}^k
$$
We need to derive the following first. Define $r = \frac{a_{N + 1}}{a_N}$. Since $\frac{a_k}{a_{k -1}} \leq \frac{a_{k + 1}}{a_k}$ for all $k \geq 1$, we have
$$
\frac{a_1}{a_0} \leq \frac{a_2}{a_1} \leq \dotsb \leq \frac{a_{N + 1}}{a_N} = r
$$
Hence
$$
a_1 \leq a_0r \\
a_2 \leq a_1r \leq a_0r^2 \\
\vdots
$$
$$
a_k \leq a_{k - 1}r \leq \dotsb \leq a_0r^k = a_0\left(\frac{a_{N + 1}}{a_N}\right)^k \tag{1}
$$
Now, the solution will be by induction. For all $k \in \{0, 1\}$, it is trivial. Now assume the fact is true for $N \geq 1$. Then we have: for any $k \in \{0, \dotsc, N\}$,
$$
{a_k}^N \leq {a_0}^{N - k}{a_N}^k
$$
Now, for any $k \in \{0, \dotsc, N + 1\}$, we need to prove
$$
{a_k}^{N + 1} \leq {a_0}^{N + 1 - k}{a_{N + 1}}^k
$$
The fact is trivial for $k = 0$ and $k = N + 1$. For $k \in \{1, \dotsc, N\}$, we can use the inductive hypothesis and $(1)$ to get,
$$
{a_k}^{N + 1} = {a_k}^Na_k \leq {a_0}^{N - k}{a_N}^ka_k \leq {a_0}^{N - k}{a_N}^k \cdot a_0\left(\frac{a_{N + 1}}{a_N}\right)^k = {a_0}^{N + 1 - k}{a_{N + 1}}^k
$$
which completes the inductive proof. I hope this inductive proof was slick enough. If $a_k = 0 $ for some $k$ then ${a_{k + 1}}^2 \leq a_ka_{k + 2} = 0 \implies a_{k + 1} = 0$ and you can see we can repeat this and so $a_j = 0$ for all $j \geq k$. Similarly we can also get $a_j = 0$ for all $j \leq k$, so all terms are $0$ and the fact becomes trivial.
A: The log-ratios of consecutive terms is increasing:
$$
a_k^2\le a_{k-1}a_{k+1}\implies
\log\left(\frac{a_k}{a_{k-1}}\right)
\le\log\left(\frac{a_{k+1}}{a_k}\right)
$$
The average of earlier log-ratios is smaller than the average of later ones. Thus, for $0\lt k\lt n$,
$$
\begin{align}
\frac1k\sum_{j=1}^k\log\left(\frac{a_j}{a_{j-1}}\right)
&\le\frac1{n{-}k}\sum_{j=k+1}^n\log\left(\frac{a_j}{a_{j-1}}\right)\\
\frac1k\log\left(\frac{a_k}{a_0}\right)
&\le\frac1{n{-}k}\log\left(\frac{a_n}{a_k}\right)\\
\left(\frac{a_k}{a_0}\right)^{n-k}
&\le\left(\frac{a_n}{a_k}\right)^k\\[6pt]
a_k^n&\le a_0^{n-k}a_n^k\\[9pt]
a_k&\le a_0^{\large1-\frac kn}a_n^{\large\frac kn}
\end{align}
$$
The cases for $k=0$ and $k=n$ are direct.
