I have a cubic univariate polynomial parameterized by rational numbers $s,t$
$$T^3+Ts + t \in \mathbb Q[T]$$
such that $t$ is in the open interval $]0,1[$ and $s$ satisfies $4s^3+27t^2 > 0$ (i.e. the polynomial has only one real root). For these conditions the real root is in fact always negative. For these ranges of $s$ and $t$ satisfying the condition on the discriminant I want to prove the following:
If $\alpha(s,t)$ be the (only) real (negative) root of this polynomial depending on $s$ and $t$ then one has
$$-\frac{\beta(s,t)}{\alpha(s,t)} - \frac{1}{2} > 0$$
where $\beta(s,t)$ is the negative real root of $$T^2 - 2(\alpha(s,t)-\alpha(s,t)^{-1}(1-s))T+1$$
It is easy to check using a computer algebra system that such an inequality with $\beta$ arises exactly because of the discriminant condition (the inequality with $s$ and $t$) but I cannot prove this at all. I thought of trying to solve the critical point of the left side of the above inequality with $\beta$ and $\alpha$ as functions of $s$ and $t$ (we know that at the defined ranges of $s$ and $t$, $\beta$ and $\alpha$ is continuous). But it all seems to be very messy to compute in order to prove this fact theoretically (i.e. because of the inequality $4s^3+27t^2>0$ I get the inequality $-\beta(s,t)/\alpha(s,t) - 1/2 > 0$). Any ideas on how to prove this? I can show the Mathematica code in how to conclude this intuitively but I do not find any clean proof for this.