In the following we use the following notation: for $a, b \in \mathbb{R}^k$, $a\cdot b:=a^Tb$ and $|a|^2:=a\cdot a$

Let $n,k\geq 1$.

Define $C(n,k)$ to be the maximum value of $C$ s.t. the following inequality holds for all $a_1, \ldots, a_n\in \mathbb{R}^k$, $b_1, \ldots, b_n\in \mathbb{R}^k$ with $a_i^T b_i = 0$ for $i=1,\dots, n$:
$$
\sum_{i=1}^n |a_i|^2 + \sum_{i=1}^n |b_i|^2 \geq \frac{C}{n} \sum_{i,j=1}^n | a_i^T b_j |.
$$

Then,

$C(n,1) = 4$ for $n$ even and $4n^2/(n^2-1)\leq C(n,1)\leq 4$ for $n$ odd.

$C(n,2)\geq 2\sqrt{2}=2.83$ and $\lim_{n\rightarrow \infty} C(n,2)\leq \pi$ as also observed by fedja.

Iosif proves in his answer that $C(n,k)\geq 4/\sqrt {3}$ which I also prove by a different argument.

**Proof:**

**$k=1$.**

$$\sum_{i,j=1}^{n}|a_ib_j|=\sum_{i=1}^{n}|a_i|\sum_{j=1}^{n}|b_j|$$ $a_ib_i=0$ for all $1\leq i\leq n$ implies that if $A,B$ are the number of non-zero $a_i, b_i$ respectively then $A+B \leq n$. Hence by Cauchy-Schwartz

$$\sum_{i,j=1}^{n}|a_ib_j|\leq\sqrt{AB}\sqrt{\sum_{i=1}^{n}|a_i|^2\sum_{j=1}^{n}|b_j|^2}$$

$$\leq (1/4) (A+B)(\sum_{i=1}^{n}|a_i|^2+\sum_{j=1}^{n}|b_j|^2)\leq \frac{n}{4} (\sum_{i=1}^{n}|a_i|^2+\sum_{j=1}^{n}|b_j|^2).$$

Thus we have $$\sum_{i=1}^{n}|a_i|^2+\sum_{j=1}^{n}|b_j|^2 \geq \frac{4}{n}\sum_{i,j=1}^{n}|a_ib_j|.$$

This proves that $C(n,1)\geq 4$.

For the upper bound for $n$ even we can take the $a_i$'s to have $n/2$ 1's and the rest 0's, swapping 1 and 0 for the $b_i$'s hence satisfying $a_ib_i=0$. A quick calculation shows that $$n (\sum_{i=1}^{n}|a_i|^2+\sum_{j=1}^{n}|b_j|^2 )/\sum_{i,j=1}^{n}|a_ib_j| = 4$$ in this case and hence that $C(n,1)\geq 4$ in this case.

For $n$ odd we take the $a_i$'s to have $(n-1)/2$ 1's and the rest 0's and again a quick calculation gives $4n^2/(n^2-1)$ for the same estimate showing that $C(n,1)\geq 4n^2/(n^2-1)$ in this case.

**For $k>1$ we need a few preliminaries:**

**Lemma**

For vectors $a,b,c \in \mathbb{R}^k$ with $a \cdot b=0$, by changing the signs of $a$, $b$ and $c$ we can arrange that $b \cdot c\geq 0$ and $a \cdot c \geq 0$.

**Proof** Clearly we can arrange that $a \cdot b$, $b \cdot c$ and $a \cdot c$ all have the same sign. if the common sign is positive we are done otherwise just change the sign of $a$ and $b$. $\blacksquare$

Consider the expression $\sum_{i,j=1}^{n}|a_i \cdot b_j|$. Since $b_j\cdot a_j=0$ we can apply the lemma to the vectors $a_i$, $b_j$ and $a_j$ . Hence by changing signs we can guarantee that $b_j\cdot a_i \geq 0$ and $a_j\cdot a_i \geq 0$ which shows that $0\leq |\measuredangle a_i b_j| \leq \pi/2$ and $0\leq |\measuredangle a_i a_j| \leq \pi/2$.

By the triangle inequality for arc lengths we have $\pi \geq |\measuredangle a_i b_j|+|\measuredangle a_i a_j|\geq \pi/2$ and thus $\pi/2 \geq |\measuredangle a_i a_j|\geq \pi/2 -|\measuredangle a_i b_j|\geq 0 $ and since $\sin$ is monotonically increasing in the range $[0, \pi/2]$ we have $1\geq \sin(|\measuredangle a_i a_j|) \geq \sin(\pi/2 -|\measuredangle a_i b_j|)=\cos(|\measuredangle a_i b_j|) \geq 0$.

Thus $|a_i \cdot b_j|=|a_i||b_j||\cos(\measuredangle a_i b_j)|\leq |a_i||b_j||\sin(\measuredangle a_i a_j)|$ and this also holds for the original vectors $a_i$, $b_j$ and $a_j$.

So $\sum_{i,j=1}^{n}|a_i \cdot b_j|\leq \sum_{i,j=1}^{n}|a_i||b_j||\sin(\measuredangle a_i a_j)|\leq \sqrt{\frac{1}{2} \sum_{i,j=1}^{n}\sin^2(\measuredangle a_i a_j)}(\sum_{i=1}^{n}|a_i|^2 + \sum_{i=1}^n |b_i|^2)$ by Cauchy's inequality.

Hence we need to upper bound
$\sum_{i,j=1}^{n}\sin^2(\measuredangle a_i a_j)=\sum_{i,j=1}^{n}u_{ij}^2$ with $u_{ij}:=\sin(\measuredangle a_i a_j)$

**$k=2$**

In this case since the points $a_i$ lie in the plane we can write $a_i=|a_i|(\cos(\theta_i),\sin(\theta_i)):=|a_i| r(\theta_i)$ giving $u_{ij}:=\sin(\theta_i-\theta_j)$.

Note that in this case we actually have an equality $\sum_{i,j=1}^{n}|a_i \cdot b_j|= \sum_{i,j=1}^{n}|a_i||b_j||\sin(\theta_i-\theta_j)|$.

Now consider the expression $R=\sum_{i,j=1}^{n}r(2\theta_i)\cdot r(2\theta_j)$. Clearly $R=|\sum_{i=1}^{n}r(2\theta_i)|^2.$ But also $R=\sum_{i,j=1}^{n}\cos(2(\theta_i-\theta_j))=\sum_{i,j=1}^{n}(1-2\sin^2(\theta_i-\theta_j))$

Therefore $$\sum_{i,j=1}^{n}\sin^2(\theta_i-\theta_j) = \sum_{i,j=1}^{n}1/2-R/2 =n^2/2-R/2.$$ Hence $$\sum_{i,j=1}^{n}\sin^2(\theta_i-\theta_j)\leq n^2/2$$ with equality iff $$\sum_{i=1}^{n}r(2\theta_i)=0$$ or where the centroid of the points $r(2\theta_i)$ is at the origin.

This gives $$\sum_{i,j=1}^{n}|a_i \cdot b_j|\leq \frac{1}{2}\sqrt{\sum_{i,j=1}^{n}\sin^2(\theta_i-\theta_j)}(\sum_{i=1}^{n}|a_i|^2 + \sum_{i=1}^n |b_i|^2)=\frac{n}{2\sqrt{2}}(\sum_{i=1}^{n}|a_i|^2 + \sum_{i=1}^n |b_i|^2)$$
implying that $C(n,2)\geq 2\sqrt{2}$.

For an upper bound we can take $a_i=r(\theta_i)$ to be uniformly distributed on the unit circle then noting as above that we have the equality $\sum_{i,j=1}^{n}|a_i \cdot b_j|= \sum_{i,j=1}^{n}|a_i||b_j||\sin(\theta_i-\theta_j)|=\sum_{i,j=1}^{n}|\sin(2\pi i/n-2\pi j/n)|$ and taking the limit as $n\rightarrow \infty$ this is equal to $$\frac{n^2}{4\pi^2}\int_0^{2\pi}\int_0^{2\pi}|\sin(x-y)|\,dx\,dy = \frac{n^2}{4\pi^2}8 \pi=\frac{2n^2}{\pi}.$$

Hence $\lim_{n\rightarrow \infty} C(n,2)\leq \pi$ as observed by fedja.

**k>2**

Here we simply observe that for any set of 3 points $\{a_i$, $a_j$, $a_k\}$, $u_{ij}^2+u_{jk}^2+u_{ki}^2$ is maximised when the points lie on a great circle with centroid at the origin. Therefore they form an equilateral triangle centred at the origin.

Hence $u_{ij}^2+u_{jk}^2+u_{ki}^2\leq 3\sin(2\pi/3)^2=3(\sqrt{3}/2)^2=9/4.$

Giving $\sum_{i,j=1}^{n} u_{ij}^2\leq 3n^2/4$ and $C(n,k)\geq 4/\sqrt {3}$ as proved by Iosif.

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