How to prove this inequality with this $\sqrt{n\left(x_{1}^{2}+x_{2}^{2}+\dots+x_{n}^{2}\right)} $

Solution 1:

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.