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?