How to prove this inequality with this $\sqrt{n\left(x_{1}^{2}+x_{2}^{2}+\dots+x_{n}^{2}\right)} $ $x_i \ge 0 \quad (i=1,2,\dots,n)$and $n \ge 2$. Prove or disprove:
$$
\sqrt{n\left(x_{1}^{2}+x_{2}^{2}+\dots+x_{n}^{2}\right)} \geq \sum_{c y c} \sqrt{x_{1}^{2}+\frac{\left((n-1) x_{1}-x_{2}-\dots-x_{n}\right)^{2}}{4 n(n-1)}}
$$
This is a hard problem proposed by Chen Shengli, an expert at mechanical theorem-proving in inequality. His program didn't work on this, so I'd like to share it with you and see if anyone can prove or disprove it (proof by computer is welcomed here).
 A: We need to prove that:
$$\sqrt{4n^2(n-1)\sum_{l=1}^nx_i^2}\geq\sum_{cyc}\sqrt{4n(n-1)x_i^2+(nx_i-\sum_{i=1}^nx_i)^2}.$$
Now, by C-S $$\left(\sum_{cyc}\sqrt{4n(n-1)x_i^2+(nx_i-\sum_{i=1}^nx_i)^2}\right)^2\leq$$
$$\leq\sum_{cyc}\frac{4n(n-1)x_i^2+\left(nx_i-\sum\limits_{i=1}^nx_i\right)^2}{nx_i+\sum\limits_{i=1}^nx_i}\sum_{i=1}^n\left(nx_i+\sum\limits_{i=1}^nx_i\right)$$
and it's enough to prove that:
$$\sum_{i=1}^n\frac{4n(n-1)x_i^2+\left(nx_i-\sum\limits_{i=1}^nx_i\right)^2}{nx_i+\sum\limits_{i=1}^nx_i}\leq\frac{2n(n-1)\sum\limits_{i=1}^nx_i^2}{
\sum\limits_{i=1}^nx_i}.$$
Now, since the last inequality is homogeneous, we can assume $\sum\limits_{i=1}^nx_i=n$ and we need to prove that
$$\sum_{i=1}^n\frac{4(n-1)x_i^2+n\left(x_i-1\right)^2}{x_i+1}\leq2(n-1)\sum\limits_{i=1}^nx_i^2$$ or $\sum\limits_{i=1}^nf(x_i)\geq0,$ where $$f(x)=2(n-1)x^2-(5n-4)x+7n-4-\frac{4(2n-1)}{x+1}.$$
But $$f''(x)=4(n-1)-\frac{8(2n-1)}{(x+1)^3}$$ and $f''(x)=0$ for
$$x=\sqrt[3]{\frac{4n-2}{n-1}}-1,$$ which gives an unique reflection point of $f$ on $[0,+\infty),$  which by the Vasc's HCF Theorem says that it's enough to prove $$\sum_{i=1}^n\frac{4(n-1)x_i^2+n\left(x_i-1\right)^2}{x_i+1}\leq2(n-1)\sum\limits_{i=1}^nx_i^2$$ for equality case of $n-1$ variables.
Can you end it now?
About HCF see here: https://www.researchgate.net/publication/257869014_An_extension_of_Jensen's_discrete_inequality_to_half_convex_functions
For $x_1=x_2=...=x_{n-1}=x$ and $x_n=n-(n-1)x$, where $0\leq x\leq\frac{n}{n-1}$,  I got
$$(x-1)^2(n-(n-1)x)((n-1)x+n-2)\geq0,$$ which is obvious.
A: Too long for a comment :
Partial answer :
Case $n=3$
It seems we have the following inequalities:
Let  $0.5\leq a,b,c\leq 1$ real numbers then $\exists d$ a real number such that $|d|\leq a,b,c$ or $d\geq 0$then :
$$3\cdot\frac{\left(a+d\right)^{2}}{\left(a+b+c+3d\right)^{2}}\left(a^{2}+b^{2}+c^{2}\right)-\left(a^{2}+\frac{\left(2a-b-c\right)^{2}}{24}\right)\geq 0$$
$$3\cdot\frac{\left(b+d\right)^{2}}{\left(a+b+c+3d\right)^{2}}\left(a^{2}+b^{2}+c^{2}\right)-\left(b^{2}+\frac{\left(2b-a-c\right)^{2}}{24}\right)\geq 0$$
And :
$$3\cdot\frac{\left(c+d\right)^{2}}{\left(a+b+c+3d\right)^{2}}\left(a^{2}+b^{2}+c^{2}\right)-\left(c^{2}+\frac{\left(2c-b-a\right)^{2}}{24}\right)\geq 0$$
Summing the inequalities gives a partial result  and the inequality is homogeneous .
It could gives some idea for the general case .
Hope it helps .
Edit 06/03/2022
For $a=b+\frac{\left(c-b\right)}{2}$ and $0.5\leq b \leq 0.75\leq c\leq 1$  it seems we can choose $d=\frac{c-a}{9}$
