For $n>1$. Expand everything. $$\iff \sum x_i^2 \ge +2(2n-1)A(A-x_n)+(2n-1)x_n^2$$Name $U=\frac{x_1+x_2+...+x_{n-1}}{n-1}$, $V=\frac{x_{n+1}+x_{n+2}+...+x_{2n-1}}{n-1}$ Then $$\iff \sum_{i<n} x_i^2+\sum_{i >n} x_i^2 \ge 2((n-1)U+(n-1)V+x_n)(n-1)(U+V-2x_n)/(2n-1)+2(n-1)x_n^2$$But $ \sum_{i<n} x_i^2+\sum_{i >n} x_i^2\ge (n-1)(U^2+V^2)$ We only need to prove $ (n-1)(U^2+V^2) \ge 2((n-1)U+(n-1)V+x_n)(n-1)(U+V-2x_n)/(2n-1)+2(n-1)x_n^2.$ $$\iff (U-V)^2\ge (4n-8)(x_n-U)(x_n-V)$$It's true after expanding and $U\le x_n\le V$

Uhhh Huhhh Goody One In the second line I put the wrong limit of $j$. That should be ,the limit of $j$ = $n+1(1)2n-1$

The problem can be rewritten as the following: $x_1 \le x_2 \le \ldots \le x_{2n-1}$ are real numbers such that $\sum_{i=1}^{2n-1}x_i=0$ Prove that $2\sum_{i=1}^{2n-1}x_i^2 \geq \sum_{i=1}^{2n-1}(x_i-x_n)^2$ WLOG we can assume that $x_n \geq 0$ $\iff \sum_{i=1}^{2n-1}x_i^2 \geq (2n-1)x_n^2$ By Cauchy-Schwarz inequality we get $\sum_{i=1}^{n-1}x_i^2 \geq \frac{1}{n-1}\left(\sum\limits_{i=1}^{n-1}x_i\right)^2$ $\iff \sum_{i=1}^{n-1}x_i^2 \geq (n-1)\left(\frac{\sum\limits_{i=1}^{n-1}x_i}{n-1}\right)^2$ Similarly,$\sum_{i=n+1}^{2n-1}x_i^2 \geq (n-1)\left(\frac{\sum\limits_{i=n+1}^{2n-1}x_i}{n-1}\right)^2$ Hence we just need to prove that $\left(\frac{\sum\limits_{i=1}^{n-1}x_i}{n-1}\right)^2+\left(\frac{\sum\limits_{i=n+1}^{2n-1}x_i}{n-1}\right)^2 \geq 2x_n^2$ again by Cauchy-Schwarz inequality we get $ 2LHS \geq \left(\frac{\sum\limits_{i=n+1}^{2n-1}x_i-\sum\limits_{i=1}^{n-1}x_i}{n-1}\right)^2=\left(\frac{2\sum\limits_{i=n+1}^{2n-1}x_i+x_n}{n-1}\right)^2$ and because $2\sum_{i=n+1}^{2n-1}x_i+x_n \geq 2(n-1)x_n$ hence the result is followed.

I don't get why $ (\frac{\sum_{i=n+1}^{2n-1}x_i-\sum_{i=1}^{n-1}x_i}{n-1})^2=(\frac{2\sum_{i=n+1}^{2n-1}x_i+x_n}{n-1})^2$

WolfusA wrote: I don't get why $ (\frac{\sum_{i=n+1}^{2n-1}x_i-\sum_{i=1}^{n-1}x_i}{n-1})^2=(\frac{2\sum_{i=n+1}^{2n-1}x_i+x_n}{n-1})^2$ Because I assume that $\sum_{i=1}^{2n-1}x_i=0$.

We will prove stronger version: For any $k\in \mathbb{Z}_+$ and real numbers $x_1 \le x_2 \le \ldots \le x_{2k-1}$ whose arithmetic mean equals $A$ holds

$$2\sum_{i=1}^{2n-1}\left( x_{i}-A\right)^2 \overset{?}{\geq} \sum_{i=1}^{2n-1}\left( x_{i}-x_{n}\right)^2$$Inequality holds for $n=1$, let $n>1$. Case $I: \ A\neq 0,$ Note that we can suppose $x_n\geq 0$. \[2\sum{x_i^2}-4A\sum{x_i}+(4n-2)A^2\overset{?}{\geq} \sum{x_i^2}-2x_n\sum{x_i}+(2n-1)x_n^2\]\[(2n-1)(2A^2-x_n^2)+\sum{x_i^2}\overset{?}{\geq} 2(2A-x_n)\sum{x_i}\]\[(2n-1)(\frac{2(\sum{x_i})^2}{(2n-1)^2}-x_n^2)+\sum{x_i^2}\overset{?}{\geq} 2(\frac{2\sum{x_i}}{n-1}-x_n)\sum{x_i}\]Since the inequality and condition are homogenous, we can assume that $\sum{x_i}=2n-1$. \[(2n-1)(2-x_n^2)+\sum{x_i^2}\overset{?}{\geq} 2(2-x_n)(2n-1)\]Let $x_1+...+x_{n-1}=p$ and $x_{n+1}+...+x_{2n-1}=q$. Set $n-1=a$ and $x_n=x$. \[4n-2-2nx^2+x^2+\sum{x_i^2}\overset{?}{\geq} 8n-4-(4n-2)x\]\[-(2n-1)x^2+\sum{x_i^2}\overset{?}{\geq} 4n-2-(4n-2)x\]Dividing both sides by $2n-1$ yields \[-x^2+\frac{\sum{x_i^2}}{2n-1}\overset{?}{\geq} 2-2x\iff \frac{\sum{x_i^2}}{2a+1}\overset{?}{\geq} (x-1)^2+1\]We have $\sum{x_i^2}=(x_1^2+...+x_{n-1}^2)+x^2+(x_{n+1}^2+...+x_{2n-1}^2)\geq \frac{p^2}{n-1}+x^2+\frac{q^2}{n-1}$. \[\frac{\frac{p^2+q^2}{a}+x^2}{2a+1}\overset{?}{\geq} (x-1)^2+1\]Note that $p\leq ax\leq q$. Let $f(p,q)=p^2+q^2$ where $p\in [0,ax]$ and $q\in [ax,2n+1-x]$, also $p+q=2n+1-x$. Let $k\leq l$, since $f(k,l)\geq f(m,k+l-m)\iff k^2+l^2\geq m^2+k^2+l^2+m^2+2kl-2km-2lm\iff m^2+kl-mk-ml\leq 0$ $\iff (m-k)(m-l)\leq 0$, $m$ must be between $k$ and $l$. Let's get the case where $f(p,q)$ is minimum where $p\leq q$. We cannot increase $p$ and decrease $q$. If $p<ax$ and $q>ax,$ then $f(p,q)\geq f(ax,2n-1-ax-x)$ which is a contradiction hence the sets satisfy $\{p,q\}=\{ax,2a+1-x-ax\}$. \[LHS(2a+1)\geq \frac{(ax)^2+(2a+1-x-ax)^2}{a}+x^2\overset{?}{\geq}(2a+1)x^2-(2a+1)2x+(4a+2)\]\[\frac{4a^2+4a+1+x^2+2ax^2+a^2x^2-(4a+2)(x+ax)}{a}\overset{?}{\geq}ax^2-(4a+2)(x-1)\]\[4a^2+4a+1+x^2+2ax^2+a^2x^2-(4a+2)(x+ax)\overset{?}{\geq} a^2x^2-(4a+2)(ax-a)\]\[4a^2+4a+1+x^2+2ax^2\overset{?}{\geq} (4a+2)(a+x)=4a^2+4ax+2a+2x\]\[2a+1+x^2+2ax^2\overset{?}{\geq} 4ax+2x\iff (x-1)^2(2a-1)\geq 0\]Which is the desired result.$\square$ Case $II: \ A=0$. Then, $\sum{x_i}=0$. $$2\sum{x_i^2}=2\sum_{i=1}^{2n-1}\left( x_{i}-A\right)^2 \overset{?}{\geq} \sum_{i=1}^{2n-1}\left( x_{i}-x_{n}\right)^2=\sum{x_i^2}+(2n-1)x_n^2$$\[\sum{x_i^2}\overset{?}{\geq} (2n-1)x_n^2\]As we proved previously, $f(x_i,x_j)\geq f(\frac{x_i+x_j}{2},\frac{x_i+x_j}{2})$ ($RHS$ doesn't get changed during this process) thus, we can suppose that $b=x_1=...=x_{n-1}$ and similarily $c=x_{n+1}=...=x_{2n-1}$. $b(n-1)+c(n-1)+x_n=0$ and $b,c\geq 0$. \[(b^2+c^2)(n-1)+x_n^2\overset{?}{\geq} (2n-1)x_n^2\iff b^2+c^2\overset{?}{\geq} 2x_n^2\]Also we can get $b,c$ closer until one of $b,c$ becomes $x_n$. If $c=x_n,$ then $b(n-1)+cn=0$. \[b^2+c^2\overset{?}{\geq} 2c^2\iff b^2\overset{?}{\geq} c^2\]Since $b\leq 0\leq c,$ we have $0=(b+c)(n-1)+c>(b+c)(n-1)\geq (b+c)$, also $b-c<0$ thus, $(b-c)(b+c)\geq 0$. If $b=x_n,$ then $c(n-1)+bn=0$. \[b^2+c^2\overset{?}{\geq} 2b^2\iff c^2\overset{?}{\geq} b^2\]$0=(b+c)(n-1)+b\leq (b+c)(n-1)$ which yields $b+c\geq 0$, also $c-b\geq 0$ hence $(c-b)(c+b)\geq 0$ As desired.$\blacksquare$