= Solution
The <Liouville approximation theorem> states that if $\alpha$ is an irrational algebraic number of degree $d\geq2$, then there is a constant $c(\alpha)>0$ such that every reduced rational $p/q$ satisfies
$$
\boxed{\left|\alpha-\frac pq\right|>
\frac{c(\alpha)}{q^d}}.
$$
Let $P\in\mathbb Z[X]$ be the <minimal polynomial> of $\alpha$, of degree $d$. Since $P$ is irreducible of degree greater than one, $P(p/q)\ne0$. Moreover,
$$
q^dP(p/q)\in\mathbb Z\setminus\{0\},
$$
and hence
$$
|P(p/q)|\geq q^{-d}.
$$
On the compact interval $I=[\alpha-1,\alpha+1]$, put
$$
M=\max_{x\in I}|P'(x)|.
$$
If $p/q\in I$, the <mean value theorem> gives a point $\xi$ between $\alpha$ and $p/q$ such that
$$
|P(p/q)|=|P'(\xi)|\left|\frac pq-\alpha\right|
\leq M\left|\frac pq-\alpha\right|.
$$
Thus $|\alpha-p/q|\geq M^{-1}q^{-d}$. If $p/q\notin I$, then $|\alpha-p/q|>1\geq q^{-d}$. Taking any
$$
0<c(\alpha)<\min\{1,M^{-1}\}
$$
proves the stated strict inequality.
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