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.

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.
