Let $G=(g_{ij}\colon i,j=1,\dots,6)$ be the $6\times6$ Gram matrix of $6$ unit vectors in $\Bbb R^3$. Let $$u:=\sum_{1\le i<j\le 6}g_{ij}^2,\quad v:=\sum_{1\le i<j<k\le 6}g_{ij}g_{ik}g_{jk}.$$
Can then $u-v$ be $<23/8$ given that $u\le15/4$?
Remark 1: A numerical experiment suggests that $23/8$ is the exact lower bound on $u-v$ given $u\le15/4$.
Remark 2: If the restriction "in $\Bbb R^3$" is removed, then, of course, one can make all the off-diagonal entries of $G$ zero.
I will show that $u\leq 15/4$ implies $23/8\leq u-v\leq 25/8$. More generally, I will show that $$u\geq 3\qquad\text{and}\qquad |u-v-3|\leq\left(\frac{u}{3}-1\right)^{3/2}.\tag{$\ast$}$$ Let $\lambda_1\geq\dotsb\geq\lambda_6\geq 0$ be the eigenvalues of $G$. The rank of $G$ is at most $3$, because the underlying unit vectors span a space of dimension at most $3$. Hence $\lambda_4=\lambda_5=\lambda_6=0$, and \begin{align*} \lambda_1+\lambda_2+\lambda_3&=\operatorname{tr}G=6,\\ \lambda_1^2+\lambda_2^2+\lambda_3^2&=\operatorname{tr}G^2=6+2u,\\ \lambda_1^3+\lambda_2^3+\lambda_3^3&=\operatorname{tr}G^3=6+6u+6v. \end{align*} Therefore, using the Newton-Girard formulae, $$\lambda_1\lambda_2+\lambda_2\lambda_3+\lambda_3\lambda_1=15-u \qquad\text{and}\qquad \lambda_1\lambda_2\lambda_3=20-4u+2v.$$ The upshot is that $$\prod_{i=1}^3(t-\lambda_i)=t^3-6t^2+(15-u)t-(20-4u+2v).$$ For prettiness we shift this polynomial by $2$: $$\prod_{i=1}^3(t+2-\lambda_i)=t^3-(u-3)t+2(u-v-3).$$ This cubic polynomial has three real roots (counted with multiplicity), whence its discrimant is nonnegative: $$(u-3)^3\geq 27(u-v-3)^2.$$ This is equivalent to $(\ast)$, and we are done.
