Consider a degree-$d$ algebraic integer $\alpha$ all of whose conjugates (including itself) are real numbers greater than 1. Its Mahler measure $M(\alpha)$ is simply equal to the norm $N(\alpha)$. Since $N(\alpha-1) \geq 1$, an application of $H\ddot{o}lder's$ inequality gives the lower bound $N(\alpha) \geq 2^d$, with equality achieved only by $\alpha = 2$. 


**Question.** Does there exist $C > 2$ such that $N(\alpha) \geq C^d$ for all $\alpha \neq 2$? 



Mahler measure of a totally positive, expanding algebraic integer
The answer is yes.


Suppose that $\alpha$ is totally real algebraic integer, and that all its conjugates are greater than $1$. Suppose also that $\alpha \ne 2$.


Note the elementary inequality for $x \in (1,\infty) \setminus \{2\}$:


$$\ln|x| \ge \frac{\ln|x-1| + \ln|x-2|}{3} + \ln C,$$


where $C = 2^{1/3} 3^{1/2} = 2.18 \ldots > 2$. (Equality holds precisely at $3 \pm \sqrt{3}$.)


Denote the conjugates of $\alpha$ by $\alpha\_i$. Assuming that $\alpha \ne 1,2$, we see that, because $\alpha - 1$ and $\alpha - 2$ are non-zero algebraic integers, we have inequalities


$$\sum\_{i=1}^{d} \ln | \alpha\_i - 1| = \ln N(\alpha -1) \ge 0,$$
$$\sum\_{i=1}^{d} \ln |\alpha\_i - 2| = \ln N(\alpha -2) \ge 0.$$


Hence we deduce that $\displaystyle{\ln N(\alpha) = \sum\_{i=1}^{d} \ln |\alpha\_i| \ge d \cdot \ln C}$, and thus $N(\alpha) \ge C^{d}$ where $C > 2$. 


