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][1] 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.

  [1]: https://en.wikipedia.org/wiki/Newton%27s_identities