Variable inequality: Show that $ {(\sum\limits_{i = 1}^{2n + 1} {{a_i}} )^2} \geqslant 4n\sum\limits_{i = 1}^{n + 1} {{a_i}{a_{i + n}}}.$ For any positive integer $n$, and real numbers (not necessarily positive)  $a_1\geqslant a_2 \geqslant …\geqslant a_{2n+1}$, show that $$
{(\sum\limits_{i = 1}^{2n + 1} {{a_i}} )^2} \geqslant 4n\sum\limits_{i = 1}^{n + 1} {{a_i}{a_{i + n}}}.$$
What I've tried: I set $x_i :=a_i - a_{i+1}$ for $i\leqslant 2n$ and $x_{2n+1}:=a_{2n+1}$, and calculate the coefficients on both sides, but gradually  find it difficult to go further, perhaps it just can't.
Please help.
Something more: if all $a_i=1$ except $a_{2n+1}=0$, the equality holds.
 A: I'm close to a solution,
but I can't go all the way,
so I'll show what I've got
in the hope that
someone else
can complete the proof.
Let
$a_i = a-b_i$,
where
$a = a_1$ and
$b_1 = 0$
so $b_i \ge 0$
and
$b_i \le b_{i+1}$.
The inequality becomes
${(\sum\limits_{i = 1}^{2n + 1} {(a-b_i)} )^2} 
\ge 4n\sum\limits_{i = 1}^{n + 1} {(a-b_i)(a-b_{i + n})}
$.
The left side is,
if
$B = \sum\limits_{i = 1}^{2n + 1} b_i$,
$\begin{array}\\
(\sum\limits_{i = 1}^{2n + 1} {(a-b_i)} )^2
&=((2n+1)a-\sum\limits_{i = 1}^{2n + 1} b_i )^2\\
&=((2n+1)a-B )^2\\
&=(2n+1)^2a^2-2(2n+1)aB+B^2\\
\end{array}
$
The right side is
$\begin{array}\\
4n\sum\limits_{i = 1}^{n + 1} {(a-b_i)(a-b_{i + n})}
&=4n\sum\limits_{i = 1}^{n + 1} (a^2-a(b_i+b_{i+n})+b_ib_{i + n})\\
&=4n((n+1)a^2-\sum\limits_{i = 1}^{n + 1}a(b_i+b_{i+n})+\sum\limits_{i = 1}^{n + 1}b_ib_{i + n})\\
&=4n(n+1)a^2-4na\sum\limits_{i = 1}^{n + 1}(b_i+b_{i+n})+4n\sum\limits_{i = 1}^{n + 1}b_ib_{i + n}\\
&=4n(n+1)a^2-4na(B+b_{n+1})+4n\sum\limits_{i = 1}^{n + 1}b_ib_{i + n}\\
&=4n(n+1)a^2-4na(B+b_{n+1})+4nS
\qquad\text{where } S=\sum\limits_{i = 1}^{n + 1}b_ib_{i + n}\\
\end{array}
$
The left-right is thus
$((2n+1)^2a^2-2(2n+1)aB+B^2)-
(4n(n+1)a^2-4na(B+b_{n+1})+4nS)\\
\quad=((2n+1)^2-4n(n+1))a^2-(2(2n+1)-4n)aB+B^2+4nab_{n+1}-4nS\\
\quad=a^2-2aB+B^2+4nab_{n+1}-4nS\\
\quad=(a-B)^2+4nab_{n+1}-4nS\\
\quad=(a-B)^2+4n(ab_{n+1}-S)\\
$
So if we can show that
$(a-B)^2+4n(ab_{n+1}-S)
\ge 0$,
or, equivalently,
$a^2-2aB+B^2+4nab_{n+1}-4nS
\ge 0$,
we are done.
At this point,
I'm stuck.
I think that
we somehow need to use
$b_i \le b_{i+1}$
to bound $S$ in relation
to $B$,
but I don't see how.
A: Inspired by marty's answer, I finally solve the problem.
Let $b_i:=a_i-b$, where $b:=a_{2n+1}$, so $b_{2n+1}=0$ and $0\leqslant b_i\leqslant b_{i+1}$ for every $i$.
The equality becomes $${\left( {\sum\limits_{i = 1}^{2n + 1} {\left( {{b_i} + b} \right)} } \right)^2} \geqslant 4n\sum\limits_{i = 1}^{n + 1} {\left( {{b_i} + b} \right)} \left( {{b_{i + n}} + b} \right)$$
$$\Leftrightarrow {\left( {\sum\limits_{i = 1}^{2n} {{b_i}}  + (2n + 1)b} \right)^2} \geqslant 4n\sum\limits_{i = 1}^{n + 1} {\left( {{b^2} + {b_i}b + {b_{i + n}}b + {b_i}{b_{i + n}}} \right)}$$
$$\Leftrightarrow (4{n^2} + 4n + 1){b^2} + (\sum\limits_{i = 1}^{2n} {{b_i}{)^2}}  + (4n + 2)b\sum\limits_{i = 1}^{2n} {{b_i}}  \geqslant (4{n^2} + 4n){b^2} + 4nb({b_{n + 1}} + \sum\limits_{i = 1}^{2n} {{b_i}} ) + 4n\sum\limits_{i = 1}^n {{b_i}{b_{i + n}}}$$
$$\Leftrightarrow {b^2} + (\sum\limits_{i = 1}^{2n} {{b_i}{)^2}}  + 2b\sum\limits_{i = 1}^{2n} {{b_i}}  \geqslant 4nb{b_{n + 1}} + 4n\sum\limits_{i = 1}^n {{b_i}{b_{i + n}}}$$
$$\Leftrightarrow {(b + \sum\limits_{i = 1}^{2n} {{b_i}}  - 2n{b_{n + 1}})^2} \geqslant 4{n^2}b_{n + 1}^2 + 4n\sum\limits_{i = 1}^n {{b_i}{b_{i + n}}}  - 4n{b_{n + 1}}\sum\limits_{i = 1}^{2n} {{b_i}}$$
$$\Leftarrow 4{n^2}b_{n + 1}^2 + 4n\sum\limits_{i = 1}^n {{b_i}{b_{i + n}}}  - 4n{b_{n + 1}}\sum\limits_{i = 1}^{2n} {{b_i}}  \leqslant 0$$
$$\Leftrightarrow \sum\limits_{i = 1}^n {({b_{n + 1}} - {b_i})({b_{n + 1}} - {b_{i + n}})}  \leqslant 0$$, which is obviously true since $${b_1} \geqslant {b_2} \geqslant  \cdots  \geqslant {b_{2n}}.$$
To complete, it's easy to find that the equality holds iff $${a_1} = {a_2} =  \cdots  = {a_{2n}} \geqslant {a_{2n + 1}} = 0.$$
A: Let's $a_{n+1} = a$ and for $i = 1,\dots,n$


*

*$a_i = a + x_i$;

*$a_{i+n+1} = a - y_{i+1}$


with both $x_i,y_i \ge 0$. Also let's $x_{n+1} = y_1 = 0$. For example, if $n=2$ we use the following notation:
$$
a_1 = a+x_1,\; a_2 = a+x_2,\; a_3 = a,\; a_4 = a-y_2,\; a_5 = a-y_3.
$$
Finally, let's $\sum x_i = X$, $\sum y_i = Y$ and $X-Y = S$.
Using this notation we have for sum in RHP:
$$
\sum_{i=1}^{n+1}a_ia_{i+n} = \sum_{i=1}^{n+1}(a + x_i)(a - y_i) \le \sum_{i=1}^{n+1}a(a+x_i-y_i) = (n+1)a^2 + aS.
$$
Now we need to prove that
$$
\left(\sum_{i=1}^{2n+1}a_i\right)^2 = \big((2n+1)a + S\big)^2 \ge 4n(n+1)a^2 + 4naS.
$$
We have
\begin{align}
\big((2n+1)a + S\big)^2 = (2n+1)^2a^2 + S^2 + 2(2n+1)Sa,
\end{align}
thus
\begin{align}
\big((2n+1)a + S\big)^2 - 4n(n+1)a^2 - 4naS = \\
a^2 + S^2 + 2aS = (a+S)^2 \ge 0.
\end{align}
QED.
