TITLE: Find the minumum of the value $k$ such two condition $x_1+x_2+ \cdots +x_k<\frac{x_1^3+x_2^3+ \cdots +x_k^3}{2}$
QUESTION [9 upvotes]: Let $x_1,x_2,\ldots, x_k$ be positive real numbers satisfying
\begin{cases} x_1+x_2+ \cdots +x_k<\frac{x_1^3+x_2^3+ \cdots +x_k^3}{2};\\ x_1^2+x_2^2+ \cdots +x_k^2<\frac{x_1+x_2+ \cdots +x_k}{2}.\end{cases} Find minimal value of $k$ satisfying those equations.
It is said this problem creat  Kvant? And the answer is $k_{\min}=516$. See here :hard inequality.
ADD it Michael Rozenberg take this problem   from the 28-th Tournament of Towns, autumn 2006, problem 7 and he say maybe it is open problem in Canada.
when $k=516$,I found this example such condition
$$x_{1}=5.169,x_{2}=x_{3}=\cdots=x_{516}=\dfrac{1}{8}$$
because
$$x^2_{1}+x^2_{2}+\cdots+x^2_{516}=5.169^2+\dfrac{515}{64}=34.765436,$$$$~~\dfrac{1}{2}(x_{1}+x_{2}+\cdots+x_{516})=\dfrac{1}{2}(5.169+\dfrac{515}{8})=34.772$$
and
$$(x_{1}+x_{2}+\cdots+x_{516})=(5.169+\dfrac{515}{8})=69.544$$
$$\dfrac{1}{2}(x^3_{1}+x^3_{2}+\cdots+x^3_{516})=\dfrac{1}{2}(5.169^3+\dfrac{515}{512})=69.557050592$$

REPLY [2 votes]: I restate the problem as follows:
Find the smallest positive integer $k$ such that there exist positive real numbers $x_1, x_2, \cdots, x_k$ satisfying
$2(x_1^2+x_2^2 + \cdots + x_k^2) < x_1 + x_2 + \cdots + x_k < \frac{1}{2}(x_1^3 + x_2^3 + \cdots + x_k^3).$
My solution:
Clearly, when $k=1, 2$, impossible. In the following, assume that $k \ge 3.$ 
Let $y_i = \frac{x_1 + x_2 + \cdots + x_k}{2(x_1^2+x_2^2 + \cdots + x_k^2)}x_i,\ \forall i$. We have $y_i > 0,\ \forall i$ and
$2(y_1^2+y_2^2 + \cdots + y_k^2) = y_1 + y_2 + \cdots + y_k < \frac{1}{2}(y_1^3 + y_2^3 + \cdots + y_k^3)$.
According to Vasc's Equal Variable Theorem [1, Corollary 1.9], there exist $0 < z_1 = z_2 = \cdots = z_{k-1} \le z_k$ such that
$$y_1 + y_2 + \cdots + y_k = z_1 + z_2 + \cdots + z_k,$$
$$y_1^2 + y_2^2 + \cdots + y_k^2 = z_1^2 + z_2^2 + \cdots + z_k^2,$$
$$\mathrm{and}\quad y_1^3 + y_2^3 + \cdots + y_k^3 \le z_1^3 + z_2^3 + \cdots + z_k^3.$$
Remark: Without using Equal Variable Theorem, actually it is not hard to prove this result.
Thus, it holds that
$2(z_1^2+z_2^2 + \cdots + z_k^2) = z_1 + z_2 + \cdots + z_k < \frac{1}{2}(z_1^3 + z_2^3 + \cdots + z_k^3)$.
By denoting $z_1 = z_2 = \cdots = z_{k-1}=b, \ z_k = a$, we have
$$0 < b \le a, \quad
2((k-1)b^2 + a^2) = (k-1)b + a < \frac{1}{2}((k-1)b^3 + a^3)$$
$$\Longleftrightarrow$$
$$b \in (0, \frac{1}{2}),\quad a = \frac{1}{4} + \frac{1}{4}\sqrt{Q}, \quad (2b(1-2b)(k-1)-7)\sqrt{Q}
> 2b(-8b^2+6b+13)(k-1) + 7$$
$$\Longleftrightarrow$$
$$b \in (0, \frac{1}{2}),\quad a = \frac{1}{4} + \frac{1}{4}\sqrt{Q}, \quad k > \frac{2(-4b^2+2b+7)(b^2-b+2)}{b(1-2b)^3}.\qquad\qquad (3)$$
where $Q = 8b(1-2b)k + (4b-1)^2$.  
Let $f(b) = \frac{2(-4b^2+2b+7)(b^2-b+2)}{b(1-2b)^3}$.
It is easy to prove that $515 < \min_{b\in (0, \frac{1}{2})} f(b) < 516$.
Thus, there exist $b\in (0, \frac{1}{2})$ and $b\le a$ satisfying (3) if and only if $k \ge 516.$
Thus, we have $k_{\min} = 516.$
Remark: When $k \ge 516$, (3) is always true by letting $b = \frac{1}{8}, \ a = \frac{1}{4} + \frac{1}{8}\sqrt{3k+1}$.
Reference
[1] Vasile Cirtoaje, “The Equal Variable Method”, J. Inequal. Pure and Appl. Math., 8(1), 2007.
Online: https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf