We prove a stronger statement, that $|px + q + \frac{8}{x}| \ge 1$ for one of $x \in \{1, 2, 4\}$. Let us write:
$$f(x) = px + q + \frac{8}{x}$$
We compare $f(2)$ to the linear interpolation of $f(1)$ and $f(4)$, which is:
$$\begin{aligned} \frac{2}{3}f(1) + \frac{1}{3}f(4) &= \left(\frac{2}{3} + \frac{4}{3}\right)p + \left(\frac{2}{3} + \frac{1}{3}\right)q + \frac{2}{3}8 + \frac{1}{3}2 \\ &= 2p + q + 6 \\ &= f(2) + 2. \end{aligned}$$
Subtracting one from each side:
$$\frac{2}{3}(f(1) - 1) + \frac{1}{3}(f(4) - 1) = f(2) + 1.$$
Now each side of these equations is either non-negative, or negative.
If the left hand side is non-negative, then:
$$\frac{2}{3}(f(1) - 1) + \frac{1}{3}(f(4) - 1) \geq 0.$$
This implies one of the terms in the average is non-negative, and so at least one of $f(1) \ge 1$ or $f(4) \ge 1$.
If on the other hand the right hand side is negative, then $f(2) < -1$.

In either case, we have found $x$ with $|f(x)| \ge 1$ as we were required to prove.

Remark: This inequality is best possible. Suppose we take $p = 2$ and $q = -9$. Then, for $1 \le x \le 4$:
$$1 - f(x) = 1 - 2x + 9 - \frac{8}{x} = \frac{2(x-1)(4-x)}{x} \ge 0.$$