find minimum value of $n$ that satisfies this inequality: $4(x_1^2+x_2^2+...+x_n^2)<2(x_1+x_2+...+x_n)I saw this question, it looks very hard:

there are real positive numbers 
  $\{x_1,x_2,...x_n\}$ , given the inequality: $$4(x_1^2+x_2^2+...+x_n^2)<2(x_1+x_2+...+x_n)<x_1^3+x_2^3+...+x_n^3$$ What is the minimum value of $n$, for this to be possible?

This is what I've tried so far:
$$\sum_{k=1}^nx_k^3>\sum_{k=1}^n4x_k^2\\\sum_{k=1}^nx_k^3-\sum_{k=1}^n4x_k^2>0\\\sum_{k=1}^nx_k^2(x_k-4)>0$$So it must be at least one number that is bigger than $4$.
So I go with trial and error .Example 1;$$x_1=10, n=3001, x_2=x_3=...=x_{3001}=0.1\\4(100+3000\cdot0.01)=520<2(10+3000\cdot0.1)=620<1000+3000\cdot0.001=1003$$ Example 2;$$x_1=6, n=1001, x_2=x_3=...=x_{1001}=0.1\\4(36+1000\cdot0.01)=184<2(6+1000\cdot0.1)=212<216+1000\cdot0.001=217$$ S0 $n$ is getting smaller, but I have no idea how to minimize $n$, or even how to approach this kind of question.
Thanks in advance for any help.
 A: This is a partial answer.  Suppose that $m$ is the smallest positive integers such that there exists $x_1,x_2,\ldots,x_m>0$ for which $$4\,\sum_{i=1}^m\,x_i^2<2\,\sum_{i=1}^m\,x_i<\sum_{i=1}^m\,x_i^3\,.$$
Without loss of generality, let $x_1\leq x_2\leq \ldots\leq x_m$.  We already know from other answers that $m\leq 516$.  Obviously, $x_1<\dfrac12$ and $x_m>4$. 
We note that
$$2x_m^2-x_m<\sum_{i=1}^{m-1}\,x_i\,\left(1-2\,x_i\right)\leq \frac{m-1}{8}\leq \frac{515}{8}\,.$$
That is,
$$x_m<\frac{1+2\sqrt{129}}{4}<5.93<6\,.$$
Now, for $t>0$, we have 
$$t(2x_m^2-x_m)-(x_m^3-2x_m)<\sum_{i=1}^{m-1}\,\Big(t(x_i-2x_i^2)-(2x_i-x_i^3)\Big)\,.\tag{*}$$
Define the function $f_t:\mathbb{R}\to\mathbb{R}$ by
$$f_t(y):=t(y-2y^2)-(2y-y^3)\,.$$
Let $\tau:=\dfrac{106}{23}$.  We note that $f_\tau$ is strictly decreasing on the interval $[4,6]$ and the maximum value of $f_\tau(y)$ for $y\in[0,6]$ is $\dfrac{61600}{328509}$, and this happens when $y=\dfrac{10}{69}$.
That is, using (*), we obtain
$$-f_\tau(4)<(m-1)\,f_\tau\left(\frac{10}{69}\right)\,,\text{ or }m>\frac{42959}{110}>390\,.$$
This proves that $m\geq 391$.  
If I could prove that $x_m>5$, then it would follow that $m\geq 494$, which is quite an improvement.  Better yet, if I knew that $x_m>5.2$, then I would get $m\geq 508$.  If I used $f_T$ with $T:=\dfrac{3}{8}+\dfrac{1519}{32\sqrt{129}}$ instead, then I would improve the three inequalities by a little:


*

*$x_m>4$ implies $m\geq 395$;

*$x_m>5$ implies $m\geq  496$;

*$x_m>5.2$ implies $m\geq 511$.


If I study $f_s$ with $s:=\dfrac{73}{19}$ instead, then I can improve the bound for $x_m>4$ by a lot.  That is,
$$x_m>4\text{ implies }m\geq 461\,.$$
This is the best bound so far without extra assumptions on $x_m$, and I have no idea how to improve it (except very slightly, like $x_m>5$ leads to $m\geq 464$, which is not a very impressive improvement using $f_s$).  Thus, so far, I can only say that $m\geq 461$, missing the intended target by $55$, but that may be improved if I can get a better bound for $x_m$.
