TITLE: A surprising inequality about a $\limsup$ for any sequence of positive numbers
QUESTION [19 upvotes]: Prove the following:

Let $\{x_n\}_{n=1}^{\infty}$ be a sequence of positive real numbers. Then
$$ \limsup \frac{x_1+x_2+...+x_n+x_{n+1}}{x_n} \geq 4 \ .$$

Note:

Yes, $4$ is sharp: $x_n=2^n$ sequence.
I have already proved that
$$ \liminf   x_n < \infty \implies \limsup \frac{x_1+x_2+...+x_n+x_{n+1}}{x_n} = \infty.$$
So, the interesting case is when $\lim x_n = \infty$.

REPLY [4 votes]: Solution(as in my lecture)
Suppose that for a positive constant $c$ and a number $N$ $$ \frac{a_1+\dots a_{n+1}}{a_n} \le c \quad \text{ for all }n \ge N $$
Let $b_n:= \sum_{i=1}^n a_n$, the inequality above is equivalent to
$$b_{n+1} \le c(b_{n}-b_{n-1}) \text{ for all }n \ge N$$
This form of the inequality is more or less more easier to deal with, but let's try it further, by letting $c_{n}$ denote $\frac{b_n}{b_{n-1}}$,then
$$c_{n+1}+\frac{c}{c_n} \le c$$
Thus $(c_n , n \ge N)$ is a sequence of bounded real numbers, thus we can choose a subsequence $(c_{n_k},k\ge 1)$ such that $$\lim_{k \rightarrow \infty} c_{n_k}= \underbrace{\limsup_{n \rightarrow \infty } c_n}_{=:a}$$
Thus
$$c \ge \limsup_{k \rightarrow \infty} \left(c_{n_k}+\frac{c}{c_{n_k-1}} \right)=a+\limsup_{k \rightarrow \infty} \frac{c}{c_{n_k-1}} \ge a+ \frac{c}{ \limsup c_n} =a+\frac{c}{a} \ge 2\sqrt{c} $$
Or, $c$ must be bigger than $4$. This means$$\text{ if} c \ge \limsup \frac{ a_1+\dots+a_{n+1}}{a_n} \text{ then } c \ge 4$$
Hence the conclusion.
$\square$
A small story: This question brings back lots of memories. The first time I encounter it was when solving the famous analysis problem books by Kaczor and Nowak. A very long time ago. Then after that, in a small tutor class with gifted highschool students, I told them that it was important to find a good presentation of your problem.
And this question was the example I used in that class.