Mahler measures of values of polynomials Let $K\ne \mathbb{Q}$ be a number field, let $\alpha\in \mathcal{O}_K$ and let $f(X)\in \mathcal{O}_K[X]$. Denote the Mahler measure by $M$.
Is there any known result about the comparison of the values $M(\alpha)$ and $M(f(\alpha))$?
 A: Let $H(\alpha)$ denote the absolute multiplicative height of $\alpha$, which is a number in $[1,\infty)$, and $M(\alpha) = H(\alpha)^d$, where $d$ is the degree of $\alpha$.  Basic properties of the height include:


*

*$H(\alpha \beta) \leq H(\alpha)H(\beta)$, for all algebraic numbers $\alpha$ and $\beta$,

*$H(\alpha_1 + \cdots + \alpha_n) \leq n \prod_{i=1}^n H(\alpha_i)$.


So if $f(x) = c_n + c_{n-1}x + \cdots + c_0x^n,$ we have
$$H(f(\alpha)) \leq (n+1)\prod_{i=0}^n H(c_i) H(\alpha)^i \leq (n+1) \left[\max_{0 \leq i \leq n} H(c_i)\right]H(\alpha)^{n(n+1)/2}.$$
Let $d$ be the degree of $\alpha$, $m = n(n+1)/2,$ and let $c = \max_{0 \leq i \leq n} H(c_i)$. Raising both sides to the $d$-th power gives
$$H(f(\alpha))^d \leq ((n+1)c)^d M(\alpha)^m.$$
Now $f(\alpha) \in \mathbb{Q}(\alpha)$, so the degree of $f(\alpha)$ is less than or equal to that of $\alpha$, and therefore $M((f(\alpha))$ is at most equal to the left-hand-side of the above inequality.  [I had made a comment similar to this, but it contained an error, and so I deleted it.]
Note that $d$ is at most $[K:\mathbb{Q}]$, and so this upper bound is of the form $CM(\alpha)^m$, where $m$ depends only on $f$ and $C$ depends on $K$ and $f$.
As for the reverse direction, I don't think any such inequality can exist, because for an arbitrary $\alpha$ (i.e. of arbitrarily large Mahler measure), we may find a polynomial such that $f(\alpha) = 1$, so $f(\alpha)$ has Mahler measure as small as possible.
A: This is all standard stuff about height functions. More generally, if we use absolute heights, then for any $f(x)\in\overline{\mathbb{Q}}[x]$ of degree $n$ there are constants $C_1(f)>0$ and $C_2(f)>0$ so that for any $\alpha\in\overline{\mathbb{Q}}$, 
$$ C_1(f)H(\alpha)^n \le H\bigl(f(\alpha)\bigr) \le C_2(f)H(\alpha)^n.$$
For $C_2(f)$ it's easy to get an explicit formula in terms of the coefficients of $f$ using the triangle inequality, as in Moreschi's answer. For $C_1(f)$, one can get an explicit formula that depends on the coefficients and the resultant of $f$. See for example Lang's Fundamentals of Diophantine Geometry.
A: This the proof of the claim in my comment to Bobby Grizzard's answer.
Since, by definition
$$ H(\alpha):=\prod_{v\in \mathfrak{S}_K} \max\{1,|\alpha|_v\}\quad (\alpha\in K), $$
one has
$$ H(\alpha)=\prod_{v\in \mathfrak{S}_{K,\infty}} \max\{1,|\alpha|_v\}\quad \forall\alpha\in \mathcal{O}_{K},$$
where $\mathfrak{S}_K$, $\mathfrak{S}_{K,\infty}$ denote the set of all places of $K$ and the set of Archimedean places of $K$ respectively.
Now, let $\alpha\in \mathcal{O}_{K}$ and $f(X)=a_n X^n+\dots+a_0\in \mathcal{O}_{K}[X]$. Then $f(\alpha)\in \mathcal{O}_K$.
Moreover, for any $v\in \mathfrak{S}_{K,\infty}$ one has
$$ |f(\alpha)|_{v}\le f_v(|\alpha|_v),\qquad f_v(X):=|a_n|_{v} X^n+\dots+|a_0|_{v}\in \mathbb{R}_{\ge 0}[X].$$
It follows that 
$$ |f(\alpha)|_{v}\le (|a_n|_{v}+\dots +|a_0|_{v}) \big(\max\{1,|\alpha|_v\}\big)^n $$
and thus
$$ \begin{split} 
H(f(\alpha))&=\prod_{v\in \mathfrak{S}_{K,\infty}} \max\{1,|f(\alpha)|_{v}\}
\\
&\le \prod_{v\in \mathfrak{S}_{K,\infty}} \max\big\{1,(|a_n|_{v}+\dots +|a_0|_{v}) \big(\max\{1,|\alpha|_v\}\big)^n\big\}
\\
& \le \prod_{v\in \mathfrak{S}_{K,\infty}} \max\{1,|a_n|_{v}+\dots +|a_0|_{v}\} \big(\max\{1,|\alpha|_v\}\big)^n
\\
&= C_f H(\alpha)^n,
\end{split} $$
where
$$ C_f:=\prod_{v\in \mathfrak{S}_{K,\infty}} \max\{1,|a_n|_{v}+\dots +|a_0|_{v}\}.$$
