Let $\alpha$ be a non-zero algebraic number of degree $d$. Denote ${\rm den}(\alpha)$ the smallest positive integer $m$ such that $m\alpha$ is an algebraic integer, and ${\rm House}(\alpha)$ the maximal absolute value among the conjugates of $\alpha$. Show that $\vert \alpha \vert \geq {\rm den}(\alpha)^{-d}\cdot{\rm House}(\alpha)^{1-d}$.
I have tried several approach but they led to nothing. For instance let $F(X)=a_0X^d+a_1X^{d-1}+…+a_{d-1}X+a_d$ be the minimal primitive polynomial in $\mathbb{Z}[X]$ that receive $\alpha$ as a root, and let $\alpha_1,…,\alpha_d$ be the conjugates of $\alpha$. Then
$${\rm den}(\alpha)^{d}\cdot{\rm House}(\alpha)^{d-1}\cdot\vert \alpha \vert \geq {\rm den}(\alpha)^{d}\cdot \prod_{i=1}^d \vert \alpha_i \vert \geq {\rm den}(\alpha)^{d}\cdot \left\vert\frac{a_n}{a_0}\right\vert,$$
but then I don't know what to do next. Another approach is that considering two case $\vert \alpha \vert < 1$ and $\vert \alpha \vert \geq 1$. Again the second case is easy but I have no idea to solve the first one. It is like the directions of the inequalities conflict and I don't know any evalution that would overcome this issue.
Any help is appreciated.
Best Answer
By multiplying the inequality by $\operatorname{den}(\alpha)$ we may assume $\operatorname{den}(\alpha)=1$, i.e. $\alpha$ is an algebraic integer. So we have to show $|\alpha|\geq\operatorname{House}(\alpha)^{1-d}$. Let $b_1,\dots,b_d$ be the absolute values of the conjugates of $\alpha$ and wlog $b_1= |\alpha|$. Let $i$ be such that $b_i=\operatorname{House}(\alpha)=\max_{j=1,\dots,d}b_j$. Then $$b_i^{d-1}b_1\geq b_1b_2\cdots b_d=|N_{\Bbb Q(\alpha)/\Bbb Q}(\alpha)|\geq1,$$ so $|\alpha|\geq\operatorname{House}(\alpha)^{1-d}$.