show this inequality with the sum $\sum_{i=1}^{n}x_{i}=n$

Question

if $n>20$,I have prove let $x_{i}>0,i=1,2,\cdots,n$,and such $$x_{1}+x_{2}+\cdots+x_{n}=n$$

show that $$F(x_{1},x_{2},\cdots,x_{n})=\left(\prod_{i=1}^{n}x_{i}\right)\cdot (\sum_{i=1}^{n}x^3_{i})\le n$$

and find all other postive integers $3\le n\le 19$

I want use $$F\left(\dfrac{x_{1}+x_{2}}{2},\dfrac{x_{1}+x_{2}}{2},x_{3},\cdots,x_{n}\right)-F(x_{1},x_{2},\cdots,x_{n})=x_{3}x_{4}\cdots x_{n}\left(\dfrac{(x_{1}+x_{2})^2}{4}\left(\dfrac{(x_{1}+x_{2})^3}{4}+\sum_{i=3}^{n}x^3_{i}\right)-x_{1}x_{2}\sum_{i=1}^nx^3_{i}\right)$$ if we show that $$\left(\dfrac{(x_{1}+x_{2})^2}{4}\left(\dfrac{(x_{1}+x_{2})^3}{4}+\sum_{i=3}^{n}x^3_{i}\right)-x_{1}x_{2}\sum_{i=1}^nx^3_{i}\right)\ge0$$ or $$\dfrac{(x_{1}+x_{2})^5}{16}-x_{1}x_{2}(x^3_{1}+x^3_{2})+\sum_{i=3}^{n}x^3_{i}\left(\dfrac{(x_{1}+x_{2})^2}{4}-x_{1}x_{2}\right)\ge0$$ or $$\dfrac{(x_{1}-x_{2})^2}{4}\left(\dfrac{1}{4}(x_{1}+x_{2})(x^2_{1}-10x_{1}x_{2}+x^2_{2})+\sum_{i=3}^{n}x^3_{i}\right)\ge 0$$

Answer 1

$$ n=2 \\ x_1 = 0.4 \\ x_2 = 1.6 $$

This meets the basic conditions.

$$ \begin{align} F() & = ( 0.4 \times 1.6 ) \times ( 0.4^3 + 1.6^3 ) \\ & = 0.64 \times ( 0.064 + 4.096 ) \\ & = 0.64 \times 4.160 \\ & = 2.66240 \\ \end{align} $$

This fails the inequality.

Note that this answer, and the comments following, predate adding the $n>20$ clause to the question.

Answer 2

Your inequality is true for any $n\geq11.$

By the Vasc's EV Method, see

https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf corollary 1.8 b),

it's enough to prove it for equality case of $n-1$ variables.

Let $x_1=x_2=...=x_{n-1}=x$ and $x_{n}=n-(n-1)x,$ where $0<x<\frac{n}{n-1}$.

Thus, we need to prove that $$x^{n-1}(n-(n-1)x)((n-1)x^3+(n-(n-1)x)^3)\leq n$$ or $f(x)\geq0,$ where $$f(x)=\ln{n}-(n-1)\ln{x}-\ln(n-(n-1)x)-\ln\left((n-1)x^3+(n-(n-1)x)^3\right),$$ which is smooth.

By the way, a proof for $n=11$ looks nice enough.