An algebra problem by Archit Tripathi

Let $f(x) = x^3 - ax + b$ and $m = \sqrt{a/3}$ . The relative maximum and minimum of $f$ are at $\left( -m , b + \frac {2a}3 m \right) \quad \text{and} \quad \left( m , b - \frac {2a}3 m \right).$ The $y$ -coordinate of the relative maximum is clearly positive. Since $f$ has three real roots, the $y$ -coordinate of the relative minimum is nonpositive, so $b - \frac {2a}3 m \leq 0 \implies b \leq \frac {2a}3 m.$ Divide both sides of the inequality on the right by $a$ to obtain $\frac ba \leq \frac 23 m \implies -m < \frac ba < \frac {3b}{2a} \leq m. \tag{1}$ Now consider the secant line of $f$ from 0 to $m$ , $y = -\frac {2a}3 x + b,$ which contains the point $\left( \frac {3b}{2a} , 0 \right).$ On the interval $(0,m]$ , $f$ is concave up, so $f$ lies on or below the secant line when $0 < x \leq m$ . This means $f \left( \frac {3b}{2a} \right) \leq 0. \tag{2}$ Additionally, $f\left(\frac ba \right) = \left( \frac ba \right)^3 > 0. \tag{3}$ By (2) and (3), $f$ goes from positive to nonpositive in the interval $\left(\frac ba , \frac {3b}{2a}\right],$ so this interval must contain a root of $f$ . By (1), this interval is a subinterval of $(-m,m]$ , which can only contain one root of $f$ , and that root necessarily has the least absolute value. Therefore, $\frac ba < p \leq \frac {3b}{2a}.$