Q:

A surprising inequality about a $\limsup$ for any sequence of positive numbers

I think I remember this from either Apostol's or Rudin's principles of Analysis. 
Claim: 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, the $4$ is attained by $x_n=2^n$ sequence. 
Try $x_n=a^n$, and this holds. 
I have already proved that UNLESS $\lim   x_n =\infty $ we will have
$$ \limsup \frac{x_1+x_2+...+x_n+x_{n+1}}{x_n} = \infty.$$
So, the interesting case is when we do have a sequence going to infinity, and it is now a matter of proving that something is invariant no matter how (exponentially, linearly, polynomially, a mix of them, etc.)  this convergence to infinity occurs. 

A:

First, assume that $\lim\sup \frac{x_{1}+\cdots+x_{n+1}}{x_{n}}=L>0$ is finite.  
Define a sequence of polynomial $\{p_{n}(x)\}$ by $p_{0}(x)=1, p_{1}(x)=x-1$ and
\begin{align}
p_{n+1}(x)=(x-1)p_{n}(x)-p_{n-1}(x)-\cdots-p_{1}(x)-p_{0}(x).\,\,(n\geq 1)
\end{align}
Claim.  $p_{n}(L)>0$ for all $n\geq 1$. 
proof.  We have $x_{1}+x_{2}+\cdots+x_{n+1}\leq Lx_{n}$ for all $n\geq 1$. 
(Edited : Actually, we only have this for sufficiently large $n$. However, this can be modified to all $n$ with a user254665's comment.) 
By induction, we will prove the following :
\begin{align}
x_{n+1}\leq \frac{p_{n}(L)}{p_{n-1}(L)}x_{n}
\end{align}
(If we prove this inequality, then claim immediately follows from $x_{n+1}, x_{n}>0$.)  Since $x_{1}+x_{2}\leq Lx_{1}$, it holds for $n=1$. If it holds for $n\leq N-1$, then 
\begin{align}
x_{N+1}&\leq (L-1)x_{N}-x_{N-1}-x_{N-2}-\cdots-x_{1} \\
&\leq \left(L-1-\frac{p_{N-2}(L)}{p_{N-1}(L)}-\frac{p_{N-2}(L)}{p_{N-1}(L)}\frac{p_{N-3}(L)}{p_{N-2}(L)}-\cdots-\frac{p_{N-2}(L)}{p_{N-1}(L)}\frac{p_{N-3}(L)}{p_{N-2}(L)}\cdots \frac{p_{0}(L)}{p_{1}(L)}\right)x_{N} \\
&=\frac{p_{N}(L)}{p_{N-1}(L)}x_{N}. 
\end{align}
Now using the recurrence relation, we can prove $p_{n+1}(x)=x(p_{n}(x)-p_{n-1}(x))$ by subtracting two recurrence relations for $n$ and $n-1$. Then by using characteristic polynomial method, we get
\begin{align}
p_{n}(x)=\frac{\sqrt{x^{2}-4x}+x-2}{2\sqrt{x^{2}-4x}}\left(\frac{x+\sqrt{x^{2}-4x}}{2}\right)^{n}+\frac{\sqrt{x^{2}-4x}-x+2}{2\sqrt{x^{2}-4x}}\left(\frac{x-\sqrt{x^{2}-4x}}{2}\right)^{n}
\end{align}
for $x\neq 4$.
Now, suppose $0<L<4$. Let
\begin{align}
\frac{L+\sqrt{L^{2}-4L}}{2}=\frac{L+i\sqrt{4L-L^{2}}}{2}=\sqrt{L}e^{i\theta}
\end{align}
where $\cos \theta=\frac{\sqrt{L}}{2}<1$. By direct computation, we have
\begin{align}
p_{n}(L)=L^{n/2}\left(\cos n\theta+\frac{L-2}{\sqrt{4L-L^{2}}}\sin n\theta\right)=\frac{2L^{n/2}}{\sqrt{4L-L^{2}}} \sin (n\theta+\alpha)
\end{align}
where $\cos\alpha=\frac{L-2}{2}$. Since $0<\theta<\pi/2$, we can choose $n$ s.t.  $\sin (n\theta+\alpha)<0\Rightarrow p_{n}(L)<0$. Contradiction.  So we have $L\geq 4$. 

There is an easier proof which is discovered by my friend.  Let $S_{n}=x_{1}+\cdots + x_{n}$ and $t_{n}=S_{n}/S_{n+1}$. Then $0<t_{n}<1$. Suppose $\limsup\frac{x_{1}+\cdots+x_{n+1}}{x_{n}}=L<4$. Then we have
\begin{align}
\frac{x_{1}+\cdots+x_{n+1}}{x_{n}}\leq L\Leftrightarrow t_{n}\geq \frac{1}{L(1-t_{n-1})}
\end{align}
for $n\geq N$ for some $N\in \mathbb{N}$. 
If we put $f(x)=\frac{1}{L(1-x)}$, $t_{n}\geq f(t_{n-1})$. We can check that $f(x)>x$ and $t_{n}$ diverges, i.e. $t_{n}>1$ for some $n$. Contradiction. 