Say $\mathbb{B}_{\ell,\infty}$ is the quaternion algebra over $\mathbb{Q}$ ramified at the finite prime $\ell$ and the infinite prime.
Say that you have two elements $\phi_1,\phi_2\in \mathbb{B}_{\ell,\infty}$ satisfying the following polynomials with integer coefficients:
$$\phi_1^2 - d_1 \phi_1 + \frac{d_1^2 - d_1}{4}=0$$ and $$\phi_2^2 - d_2 \phi_2 + \frac{d_2^2 - d_2}{4}=0$$ for some integers $d_1,d_2<-1$.
Then, their product, $\phi_1\phi_2$, satisfies some polynomial
$$(\phi_1\phi_2)^2 - t \phi_1\phi_2 + \frac{(d_1^2-d_1)(d_2^2-d_2)}{16}=0$$ of negative discriminant.
Define $m:=\frac{d_1d_2-(d_1d_2-2t)^2}{4}$; then, one can show the suborder of $\mathbb{B}_{\ell,\infty}$ generated by $1,\phi_1,\phi_2$ and $\phi_1\phi_2$ has discriminant $m^2$. I would like to say that $m \geq 0$, but I'm having trouble... (this fact is stated in passing on page 22, line 13 of this paper: https://arxiv.org/pdf/1206.6942.pdf).
Solving for our desired bounds on $t$ we want to show $$\frac{d_1d_2-\sqrt{d_1d_2}}{2}\leq t\leq \frac{d_1d_2+\sqrt{d_1d_2}}{2}.$$ Notice that this lower bound is positive. The fact that the discriminant of the minimal polynomial of $\phi_1\phi_2$ is negative is only enough to give an upper bound, and so that isn't enough. Is there some useful fact of quaternion algebras that I'm overlooking?
Thanks in advance!
The trick is to choose the representation of $\mathbb{B}_{\ell,\infty}$ so that the embedding of one of the $\mathbb{Z}[\frac{d_i+\sqrt{d_i}}{2}] = \mathbb{Z}[\phi_i]$ is particularly nice.
For example, choose a representation of $\mathbb{B}_{\ell,\infty}\hookrightarrow M_2\left(\mathbb{Q}(\sqrt{d_1})\right)$ so that the composition with $\mathbb{Z}[\phi_i]\hookrightarrow \mathbb{B}_{\ell,\infty}$ sends $$\alpha = a_0 + a_1\sqrt{d_1}\mapsto \begin{pmatrix}a_0 + a_1\sqrt{d_1} & 0\\ 0 & a_0 - a_1\sqrt{d_1}\end{pmatrix}.$$ Then, your $\phi_1$, which satisfies $\phi_1^2 - d_1\phi_1 + \frac{d_1^2 -d_2}{4}$, will map to $$\phi_1 \mapsto \begin{pmatrix}\frac{d_1+\sqrt{d_1}}{2} & 0\\ 0 & \frac{d_1-\sqrt{d_1}}{2}\end{pmatrix}.$$
Now, under this representation the element $\phi_2$ maps somewhere; say, $$\phi_2\mapsto \begin{pmatrix}\gamma & \delta\\ j^2\bar\delta & \bar\gamma\end{pmatrix}.$$
Since you know the trace of $\phi_2$ is $d_2$ you can actually rewrite this as
$$\phi_2\mapsto \begin{pmatrix}\frac{d_2}{2} + c\sqrt{d_1} & \delta\\ j^2\bar\delta & \frac{d_2}{2} - c\sqrt{d_1}\end{pmatrix}.$$
Now, you can compute the trace of the product to be $t = cd_1 + d_1d_2/2$, and similarly, you can recompute $m = (d_1d_2-(d_1d_2-2t)^2)/4 = -4c^2d_1^2 + d_1d_2$.
Now your desired inequality of $m\geq 0$ becomes $$|c|\leq \sqrt{\frac{d_2}{4d_1}}.$$
Since the norm of $\phi_2$ is just the determininant, which equals $$\det(\phi_2) = N(\gamma)-j^2N(\delta) = \frac{d_2^2-d_2}{4},$$ and $-j^2N(\delta) \geq 0$ (You can choose $j^2 = -\ell q$ or $-q$ for some rational prime $q$ depending on whether $\ell$ is inert or ramified in $\mathbb{Q}(\sqrt{d_1})$, for instance.), you find
$$N(\gamma) = \frac{d_2^2}{4} - c^2d_1\leq \frac{d_2^2-d_2}{4}.$$
Solving for $c$ gives the desired inequality.