Alternative proof of Liouville's approximation theorem

Question

Let $\alpha\in\mathbb{R}$ be an algebraic number of degree $d\geq2$. I am asked to prove Liouville's approximation theorem using the fact that $$ \mathop{\text{den}}(\alpha)^d \alpha^{(1)}\dotsm\alpha^{(d)}\quad \text{and} \quad \lvert\alpha\rvert \geq \mathop{\text{den}}(\alpha)^{-d}\,h(\alpha)^{1-d} $$ where $\mathop{\text{den}}(\alpha)$ is the smallest $m\in\mathbb{Z}_{>0}$ such that $m\alpha$ is an algebraic integer and $h(\alpha)=\max_{1\leq i\leq d}\lvert\alpha^{(i)}\rvert$ with $\alpha^{(i)}$ the $i$-th conjugate of $\alpha$.

Precisely, I need to prove that there is a constant $c(\alpha)>0$ such that $$ \left\lvert\alpha - \frac{x}{y}\right\rvert \geq c(\alpha)y^{-d} $$ for all $x,y\in\mathbb{Z}$ with $y>0$.

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So far I defined $\beta=\alpha-\frac{x}{y}$, which is again an algebraic real number of degree $d$. Hence $$ \left\lvert\alpha - \frac{x}{y}\right\rvert = \lvert\beta\rvert \geq \mathop{\text{den}}(\beta)^{-d}\,h(\beta)^{1-d} \geq y^{-d}\mathop{\text{den}}(\alpha)^{-d}\,h(\beta)^{1-d} $$ because $\mathop{\text{den}}(\beta)=\mathop{\text{den}}(\alpha-\frac{x}{y})\leq y\mathop{\text{den}}(\alpha)$. Now I'm stuck, though, because I can't find any useful bounds on $h(\beta)$.

Probably the hypothesis that $\alpha\in\mathbb{R}$ is necessary, but I can't see how it could come handy.

Answer 1

Indeed, you are not too far from the solution. Observe that if $\lvert\beta\rvert\geq1$ then any $c(\alpha)<1$ will work because $y\geq1$. Hence we can safely assume $\lvert\beta\rvert<1$, which implies $$ 1>\left\lvert\alpha-\frac{x}{y}\right\rvert\geq\left\lvert\frac{x}{y}\right\rvert-\lvert\alpha\rvert $$ therefore $$ \lvert\beta\rvert \geq \mathop{\text{den}}(\beta)^{-d}\,h(\beta)^{1-d} \geq y^{-d}\mathop{\text{den}}(\alpha)^{-d} \,(h(\alpha)+\lvert\alpha\rvert+1)^{1-d} $$ Then you can just define $c(\alpha)=\mathop{\text{den}}(\alpha)^{-d} \,(h(\alpha)+\lvert\alpha\rvert+1)^{1-d}$ because $d\geq2$ implies $c(\alpha)<1$.