TITLE: Smallest possible sum $\sum_{i=0}^{\infty} \frac{a_i^2}{a_{i+1}}$ over non-increasing infinite sequences starting with 1
QUESTION [5 upvotes]: Let $a_n$ be a sequence of positive numbers such that $a_0=1$ and $\forall n : a_n \ge a_{n+1}$.
Find the infinum of $\sum_{i=0}^{\infty} \frac{a_i^2}{a_{i+1}}$ over all such sequences.
If $\{a_n\}$ is a geometric seriess $1, q, q^2, ...$, where $0<q<1$, then the sum equals $\frac{1}{q}+\frac{1}{1-q}$, which has minimum at $q=\frac{1}{2}$, so the infinum is no greater than $4$.
I suspect that the answer is indeed $4$, but I have no idea how to prove that.

REPLY [2 votes]: The infimum is indeed $4$. 
We have
$$
\begin{align}
\sum_{i=0}^{\infty} \frac{a_i^2}{a_{i+1}}&\geq \sum_{i=0}^n \frac{a_i^2}{a_{i+1}}+\sum_{i=n+1}^{\infty} a_i \ \ \ \ \textrm{( by $a_i\geq a_{i+1}$: replace denominator.)}\\
&>\sum_{i=0}^n \frac{a_i^2}{a_{i+1}} + a_{n+1} \ \ \ \ \textrm{(drop all terms with $i\geq n+2$.)}\\
&\geq \sum_{i=0}^{n-1} \frac{a_i^2}{a_{i+1}}+2a_n \ \ \ \ \textrm{(apply AM-GM to the last two terms)}\\
&\geq \sum_{i=0}^{n-2} \frac{a_i^2}{a_{i+1}}+2\sqrt 2 a_{n-1} \ \ \textrm{(apply AM-GM to the last two)}
\end{align}
$$
We repeat this process until we obtain 
$$
\sum_{i=0}^{\infty} \frac{a_i^2}{a_{i+1}} > \frac1{a_1}+2^{1+\frac12+\cdots + \frac1{2^{n-1}}}a_{n-(n-1)}.
$$
Then applying AM-GM to this one, we obtain 
$$
\sum_{i=0}^{\infty} \frac{a_i^2}{a_{i+1}} > 2^{1+\frac12+\cdots+\frac1{2^n}}
$$
This lower bound can be made arbitrary close to $4$. Hence, the claim is proved.