Mahler measure of a totally positive, expanding algebraic integer 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 $\alpha \neq 2$? 
 A: 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 $C > 2$. 
A: Gypsum's argument is really nice. In the same spirit, we have the following inequality:
$$\ln|x| \geq \frac{2\ln|x-1| + \ln|x-2|}{5} + \ln\sqrt{5}$$
which is achieved at $x = \frac{5\pm \sqrt{5}}{2}$. This way we obtain the optimal constant $C = \sqrt{5}$. Perhaps some further tweaks can yield even larger $C$ with finitely many exceptions, as in the work of Smyth and Flammang. 
I wonder how far this approach can be extended. Smyth showed that for totally positive algebraic integers (whose conjugates are not necessarily greater than 1), their $M(\alpha)^{\frac{1}{d}}$ are dense beyond 1.73, which is very close to the lower bound cited by @BobbyGrizzard. Do we have a similar situation here? As a first step, it would be good to construct infinitely many $\alpha$ that give upper bound to $C$. For example, consider the $n$-th $Chebyshev$ polynomial $T_n(x)$. Then the monic polynomial $(-x)^nT_n(\frac{2}{x}-1)$ has all its roots greater than 1. In this case $N(\alpha) = 2^{2n-1}$, suggesting $C \leq 4$. 
