Proving: $2n (\sum_{i=1}^{2n} a_i^2) \geq (\sum_{i=1}^{2n} a_i)^2+(\sum_{i=1}^{2n} a_i (-1)^i)^2$

Question

Let $a_i\in \mathbb{R}$ and $n$ an integer.

How do you prove:

$$2n \left(\sum_{i=1}^{2n} a_i^2\right) \geq \left(\sum_{i=1}^{2n} a_i\right)^2+\left(\sum_{i=1}^{2n} a_i (-1)^i\right)^2?$$

This is how far I have got. I said let V be a inner product space with orthonormal basis $\{u_1, \dots ,u_{2n}\}$ and let, $u,v \in V$ and $u= \sum_{i=1}^{2n}a_iu_i$ and $v= \sum_{i=1}^{2n}a_iu_i$.

Therefore as

$$\langle u,v \rangle = \sum_{i=1}^{2n} a_ib_i,\quad \langle u,u \rangle = \sum_{i=1}^{2n} a_i^2,\quad \langle v,v \rangle = \sum_{i=1}^{2n} b_i^2,$$ and from the Cauchy–Schwarz Inequality $|\langle u,v \rangle|^2 \leq \langle u,u \rangle\langle v,v \rangle$ we know that

$$\left(\sum_{i=1}^{2n} a_i^2\right)\left(\sum_{i=1}^{2n} b_i^2\right) \geq \left(\sum_{i=1}^{2n} a_ib_i\right)^2$$

So letting $b_i=1$ for all $i$ we see

$$2n\left(\sum_{i=1}^{2n} a_i^2\right) \geq \left(\sum_{i=1}^{2n} a_i\right)^2$$

Clearly though $(\sum_{i=1}^{2n} a_i (-1)^i)^2 \geq 0$ so we can't simply move on from this step, I feel I need to back up a bit to continue, but i've got a bit lost, if anyone could help it'd be great!

EDIT: Ok, so let $b_i=(-1)^i$ then

$$\left(\sum_{i=1}^{2n} a_i^2\right)\left(\sum_{i=1}^{2n} ((-1)^2)^i\right) \geq \left(\sum_{i=1}^{2n} a_i(-1)^i\right)^2$$ $$\Rightarrow 2n\left(\sum_{i=1}^{2n} a_i^2\right) \geq \left(\sum_{i=1}^{2n} a_i(-1)^i\right)^2$$ $$\Rightarrow 4n\left(\sum_{i=1}^{2n} a_i^2\right) \geq \left(\sum_{i=1}^{2n} a_i\right)^2 + \left(\sum_{i=1}^{2n} a_i(-1)^i\right)^2.$$

Maybe the upper indices on the summation signs for $2n$ are typos in the question? Maybe they were meant to be $n$, then this holds?

Answer 1

The main idea is to collect terms according to their signs. Let $E := \sum \limits_{i \text{ even}} a_i$ and $O := \sum \limits_{i \text{ odd}} a_i,$ where $1 \leqslant i \leqslant 2n$. (For brevity, we omit the range of $i$ in the summation notation.) Then $\sum \limits_i a_i = E+ O$ and $\sum \limits_i (-1)^ia_i = E - O$. Using this observation,

$$ \begin{eqnarray*} \text{RHS} &=& (E+O)^2 + (E-O)^2 \\ &=& 2(E^2 + O^2) \\ &\stackrel{\color{Red}{(\ast)}}{\leqslant}& 2 \left(n\sum_{i \text{ even}} a_i^2 + n\sum_{i \text{ odd}} a_i^2 \right) \\ &=& 2n \sum_{i=1}^{2n} a_i^2 , \end{eqnarray*} $$ where $\color{Red}{(\ast)}$ follows by applying Cauchy-Schwarz inequality to the two sets of numbers $\{ a_1, a_3, \ldots, a_{2n-1} \}$ and $\{ a_2, a_4, \ldots, a_{2n}\}$.

EDIT: Condition for equality. For equality to occur, we must have equality in both the applications of Cauchy-Schwarz in $\color{Red}{(\ast)}$. This is possible if and only if $a_1 = a_3 = a_5 = \cdots = a_{2n-1}$ and $a_2 = a_4 = \cdots = a_{2n-2} = a_{2n}$; i.e., iff there exist $c_0$ and $c_1$ such that $a_i = c_{i\ \bmod 2}$ for all $i$.

Answer 2

Here is an expansion of my comment, since it gives a different solution than Srivatsan's.

Let $A=(a_i), B=(b_i),C=(c_i)$ with $b_i=1,c_i=(-1)^n$. Then $\|B\|^2=\|C\|^2=2n$, and $B\cdot C = 0$. Let $B'=B/\|B\|$ and $C'=C/\|C\|$, and let $V$ be the vector space spanned by $B$ and $C$. Then we have the following inequalities:

$$\sum a_i^2 = \|A\|^2 \geq \|\operatorname{proj}_V(A)\|^2 = \| (A\cdot B')B' + (A\cdot C')C' \|^2 = (A\cdot B')^2 + (A\cdot C')^2.$$

Multiplying both sides by $2n= \|B\|^2=\|C\|^2$ yields the result. Also, note that we have equality when $A=\operatorname{proj}_V(A)$, that is, when $A\in \operatorname{span}(B,C)$.