Irish Mathematical Olympiad

By letting $s = |t|$ we see that it is enough to show $|1 - 4s^2 - 3s| \le \frac{3}{2}$, for $0 \le s \le 1$. Now $|1 - 4s^2 - 3s| = |f(s)|$, where
$$\begin{aligned} f(s) &= |1 - 4s^2| - 3s \\ &= \begin{cases} 1 - 4s^2 - 3s, & \text{if } 0 \le s \le \frac{1}{2} \\ 4s^2 - 3s - 1, & \text{if } \frac{1}{2} \le s \le 1 \end{cases} \\ &= \begin{cases} (1+s)(1-4s), & \text{if } 0 \le s \le \frac{1}{2} \\ (4s+1)(s-1), & \text{if } \frac{1}{2} \le s \le 1 \end{cases} \\ &= \begin{cases} p(s), & \text{if } 0 \le s \le \frac{1}{2} \\ q(s), & \text{if } \frac{1}{2} \le s \le 1. \end{cases} \end{aligned}$$

Clearly, $p, q$ are quadratic functions; $q(s) = -p(-s)$; $p$ cuts crosses the horizontal axis at $-1$ and $1/4$ and is concave, while $q$ is convex and crosses the horizontal axis at $-1/4$ and $1$. Also, $p$ is decreasing on $(0, 1/2)$, negative on $(1/4, 1/2)$, while $q$ is increasing and negative on $(1/2, 1)$. Thus the graph of $f$ is a union of two parabolic arcs, which meet at $s = 1/2$ where $f$ takes its least value, namely $-3/2$. It takes its largest value at $0$. In other words,
$$-\frac{3}{2} \le f(s) \le 1, \quad 0 \le s \le 1$$
and so $|f(s)| \le \frac{3}{2}$, with equality when $s = 1/2$.