Problem 1488

1. We start with the given conditions: $p$ is a positive integer, $q$ and $z$ are real numbers such that $0 \le q \le 1$ and $q^{p+1} \le z \le 1$.
2. We need to prove the inequality:
$$\prod_{k=1}^p \left|\frac{z - q^k}{z + q^k}\right| \le \prod_{k=1}^p \left|\frac{1 - q^k}{1 + q^k}\right|.$$
3. Without loss of generality, assume $p + 1$ is the smallest positive integer such that $0 \le q^{p + 1} \le z \le 1$. This implies $0 \le q^{p+1} \le z \le q^p \le 1$.
4. We claim that:
$$\left|\frac{q^t - z}{z + q^t}\right| \le \left|\frac{1 - q^{p + 1 - t}}{1 + q^{p + 1 - t}}\right|$$
for $1 \le t \le p$.

5. To prove this claim, consider the inequality:
$$\left|\frac{q^t - z}{z + q^t}\right| \le \left|\frac{1 - q^{p + 1 - t}}{1 + q^{p + 1 - t}}\right|.$$
This inequality can be rewritten as:
$$\left|\frac{q^t - z}{z + q^t}\right| \le \left|\frac{1 - q^{p + 1 - t}}{1 + q^{p + 1 - t}}\right| \iff 0 \le 2q^t(1 - z).$$
Since $0 \le q \le 1$ and $0 \le z \le 1$, the inequality $0 \le 2q^t(1 - z)$ holds true.

6. Applying the claim, we get:
$$\prod_{k=1}^p \left|\frac{q^k - z}{z + q^k}\right| \le \prod_{k=1}^p \left|\frac{1 - q^{p + 1 - k}}{1 + q^{p + 1 - k}}\right|.$$
7. Notice that the right-hand side of the inequality can be rewritten as:
$$\prod_{k=1}^p \left|\frac{1 - q^{p + 1 - k}}{1 + q^{p + 1 - k}}\right| = \prod_{k=1}^p \left|\frac{1 - q^k}{1 + q^k}\right|.$$
8. Therefore, we have:
$$\prod_{k=1}^p \left|\frac{q^k - z}{z + q^k}\right| \le \prod_{k=1}^p \left|\frac{1 - q^k}{1 + q^k}\right|.$$

$\blacksquare$