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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.

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  • $\begingroup$ If all $a_i =1$, the inequality is $4n^2+4n+1 \ge 4n^2+4n$. Impressively close. $\endgroup$ – marty cohen Feb 24 '18 at 3:23
  • $\begingroup$ @martycohen Thanks for reminding me, I've added more details. $\endgroup$ – Juggler Feb 24 '18 at 3:32
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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.

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  • $\begingroup$ at this point, looking at that thing as a quadratic polynomial in "a" and trying to show that the discriminant is non-positive could be the way to go. $\endgroup$ – dezdichado Feb 24 '18 at 4:25
  • $\begingroup$ It seems a good thought, but something can be improved. Note that the equality holds when all $a_i=1$ except $a_{2n+1}=0$. Regarding this, it might be better if you set $b_i=a_i-a_{2n+1}$. $\endgroup$ – Juggler Feb 24 '18 at 6:02
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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 $${a_1} = {a_2} =  \cdots  = {a_{2n}} \geqslant {a_{2n + 1}} = 0.$$

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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.

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