The numbers $a = (k-1)k$, $b = (k+1)k$, $c = (k-1)(k+1)$, $d = k^2$ clearly satisfy the equality $ab = cd$ and the inequalities $a < c < d < b$ for any $k > 1$. Let thus $k$ be the least natural number for which $p < a$, i.e. $p < (k-1)k$ (for a given $p$). We will show that for this $k$ necessarily $b = (k+1)k \le p + 4 + 2\sqrt{4p+1}$, which is evidently a number by $\frac{1}{4}$ smaller than the upper bound of the interval in our problem, so we will be done.

In view of the choice of the number $k$ we have $p \ge (k-2)(k-1)$. Solving this quadratic inequality yields the estimate
$$k \le \frac{3}{2} + \sqrt{p + \frac{1}{4}},$$
from which it already follows that
$$b = (k+1)k \le \left(\frac{5}{2} + \sqrt{p + \frac{1}{4}}\right) \cdot \left(\frac{3}{2} + \sqrt{p + \frac{1}{4}}\right) \\ = \frac{15}{4} + 4\sqrt{p + \frac{1}{4}} + \left(p + \frac{1}{4}\right) = p + 4 + 2\sqrt{4p+1}.$$