Lemma: There is a number $c' \in (0,c)$ such that:
$$\liminf_{n \rightarrow \infty} \frac{ p_1+p_2+...+p_n}{H_{n+1}^2-H_n^2} \ge c',$$
where $H_{n}:=F_1+F_2+...+F_n.$
Remark The condition $c<c'$ is meaningless in the statement of the above lemma. However, we require $c'<c$ so that the later-defined function $g$ will be strictly increasing.
Proof for Lemma ( presented at the end of this post).
Solution for the main problem :
Because $p_n \ge 0$, $(x_n)$ is increasing.
Because $$x_{n+1} = x_n+\frac{2p_n}{x_1+x_2+...+x_n}\le x_n+\frac{2p_n}{nx_1},$$ by induction, we imply that : $x_{n+1}^2 \le \frac{2}{x_1}F_{n+1}^2$. Therefore, $\limsup \frac{x_n}{F_n} <\infty.$
Because $$x_{n+1}^2 = x_n^2+\frac{2p_nx_n}{x_1+x_2+...+x_n}+\frac{p_n^2}{(x_1+x_2+...+x_n)^2} \ge x_n^2+\frac{2p_n}{n},$$ by induction, we imply that : $x_{n+1}^2 \ge 2F_{n+1}^2$. Therefore, $\liminf \frac{x_n}{F_n} \ge \sqrt{2}.$
Now, let $s_n=x_1+x_2+...+x_n$ and $a \in (0, \sqrt{2c}]$ is a real number such that: $$ \liminf \frac{x_n}{F_n} \ge a $$
Now, we consider the following relation:
\begin{align}
s_{n+1}^2-2s_n^2+s_{n-1}^2&= 2s_n(x_{n+1}-x_n)+x_n^2+x_{n+1}^2=2p_n+x_n^2+x_{n+1}^2
\end{align}
Therefore, by definition of $a$ and the properties of $\liminf$, we deduce that:
$$ \liminf \frac{s_{n+1}^2-2s_n^2+s_{n-1}^2}{\frac{2}{a^2}p_n+F_n^2+F_{n+1}^2} \ge a^2. $$
Thus, by Cesaro's theorem, we have that:
$$ \liminf \frac{s_{n+1}^2-s_n^2}{S_{n+1}} \ge a^2. $$
where \begin{align}
S_{n+1}&:=\sum_{m=1}^n \left( \frac{2}{a^2}p_m+F_m^2+F_{m+1}^2 \right)\\
\end{align}
Besides,\begin{align}
H_{n+1}^2-2H_n^2+H_{n-1}^2&= 2H_n(F_{n+1}-F_n)+F_n^2+F_{n+1}^2 \\
&\sim \frac{1}{c}n(F_{n+1}+F_n)(F_{n+1}-F_n)+F_n^2+F_{n+1}^2
\\
&=\frac{p_n}{c}+F_n^2+F_{n+1}^2.
\end{align}
Thus, by combining with our lemma, we have that:
$$ \liminf \frac{1}{\left(\frac{2}{a^2}-\frac{1}{c} \right)c'+1}.\frac{s_{n+1}^2-s_n^2}{H_{n+1}^2-H_n^2} \ge a^2.$$
In consequences, $\liminf \frac{s_{n+1}^2}{H_{n+1}^2} \ge a^2\left[ \left(\frac{2}{a^2}-\frac{1}{c} \right)c'+1 \right]=: g(a).$
Therefore, $$\limsup_{n \rightarrow \infty}\frac{x_{n+1}-x_n}{F_{n+1}-F_n}=\limsup_{n \rightarrow \infty} \frac{p_n}{ s_n(F_{n+1}-F_n)}=\limsup_{n \rightarrow \infty} \frac{n(F_{n+1}+F_n)}{ s_n} \le \frac{2c}{\sqrt{g(a)}}.$$
So, we have proven that if $\liminf\frac{x_n}{F_n}> a \in (0, \sqrt{2c}]$, $\limsup \frac{x_{n}}{F_n} \le \frac{2c}{\sqrt{g(a)}}.$($*$)
Similarly, we can prove that if $\limsup \frac{x_{n}}{F_n} \le b \in [\sqrt{2c},\infty)$, then $\liminf \frac{x_{n}}{F_n} \ge \frac{2c}{\sqrt{g(b)}}.$($**$)
Let $a'= \liminf \frac{x_n}{F_n}$, i.e.
If $a'< \sqrt{2c}$, then $g(a')>(a')^2$, then by (*), $\limsup \frac{x_n}{F_n} \le \frac{2c}{a'}$. Thus, by $(**)$, $\liminf \frac{x_n}{F_n} \ge \frac{2c}{ \sqrt{g(\frac{2c}{a'})}}$. In other words, $a' \ge \frac{2c}{ \sqrt{g(\frac{2c}{a'})}}$,which is a contradiction because $0<a' <\sqrt{2c}$.
Therefore, $\liminf \frac{x_n}{F_n} \ge \sqrt{2c}$.
Similarly, we can also show that $\limsup \frac{x_n}{F_n} \le \sqrt{2c}$. Therefore, we have the desired conclusion. $\square$.
Proof for Lemma
We have:
\begin{align}
p_1+p_2+...+p_n&=nF_{n+1}^2-F_n^2-F_{n}^2-...-F_1^2\\
& \ge nF_{n+1}^2- F_{n+1}(F_1+...+F_n)=nF{n+1}^2\left(1- \frac{F_1+...F_n}{nF_n} \right)\\
&\sim \frac{c}{2}\left(2F_1+...+2F_n+F_{n+1}\right)F_{n+1}(1-c)\\
&= \frac{c(1-c)}{2}(H_{n+1}^2-H_n^2).
\end{align}
Hence, the conclusion.
$\square$.
