Let $\alpha$ be a real algebraic number of degree $\geq 3$. Prove that the following three assertions are equivalent:
- for every $\kappa >2$ there is a constant $c(\alpha,\kappa)>0$ such that $$|\xi-\alpha| \geq c(\alpha,\kappa) H(\xi)^{-\kappa}$$ for every $\xi \in \mathbb{Q}$.
- for every $\kappa >2$, the inequality $$|\xi -\alpha| \leq H(\xi)^{-\kappa}$$ in $\xi \in \mathbb{Q}$ has only finitely many solutions
I have managed $(1) \Rightarrow (2)$: Choose $\kappa'$ such that $\kappa > \kappa' > 0$
$$H(\xi)^{-\kappa} \geq |\xi -\alpha| \geq c(\alpha,\kappa')H(\xi)^{-\kappa'}$$ Implies $\dfrac{1}{c(\alpha,\kappa')} \geq H(\xi)^{\kappa-\kappa'}$. Hence $H(\xi)$ is bounded, so there are finitely many $\xi$.
From $(2)$ we have there are infinitely many $\xi$ with $ |\xi -\alpha| \geq H(\xi)^{-\kappa}$ however I can't indicate $c(\alpha,\kappa)$ such that for all $\xi$ this inequality holds.
Let $\xi_1, \ldots, \xi_\ell$ denote the finitely many solutions to the inequality in $(2)$. Note that since $\alpha$ has degree at least $3$, $\alpha \neq 0$ and thus $\xi_i \neq 0$ for any of the finite number of solutions to the inequality.
Define the constant $c$ by $$ c = \inf_{1 \leq i \leq \ell} \Big(\lvert \xi_i - \alpha \rvert H(\xi_i)^\kappa\Big).$$ As there are only finitely many terms, this $c$ is well-defined and $c > 0$. Further, we can assume $c < 1$, as otherwise there would be no exceptional $\xi_i$.
Then we necessarily have that $\lvert \xi_i - \alpha\rvert \geq c H(\xi_i)^{-\kappa}$ for each exceptional $\xi_i$. For nonexceptional $\xi_i$, the stronger inequality $\lvert \xi - \alpha\rvert \geq H(\xi)^{-\kappa} \geq c H(\xi)^{-\kappa}$ holds. Thus the inequality in $(1)$ holds for all $\xi$.