A hard VMO problem #2

If the line $d$ has the equation $y = mx + b$ , then the three intersection points $x_1$ , $x_2$ , and $x_3$ with $y = \sqrt[3]{x^2}$ are solutions to the equation $mx + b = \sqrt[3]{x^2}$ or $x^3 + \frac{3bm^2 - 1}{m^3}x^2 + \frac{3b^2}{m^2}x + \frac{b^3}{m^3} = 0$ , which by Vieta means that $x_1x_2x_3 = \frac{b^3}{m^3}$ and $x_1 + x_2 + x_3 = \frac{3bm^2 - 1}{m^3}$ .

Since $B = \sqrt[3]{\frac{x_1^2}{x_2x_3}} + \sqrt[3]{\frac{x_2^2}{x_3x_1}} + \sqrt[3]{\frac{x_3^2}{x_1x_2}}$ $=$ $\frac{1}{\sqrt[3]{x_1x_2x_3}}(x_1 + x_2 + x_3)$ , after substitution $B = \sqrt[3]{\frac{m^3}{b^3}}(\frac{3bm^2 - 1}{m^3})$ or $B = 3 - \frac{1}{bm^2}$ .

The derivative of $y = \sqrt[3]{x^2}$ at $x = p$ is $m = y' = \frac{2}{3\sqrt[3]{p}}$ , and as the tangent line of $y = \sqrt[3]{x^2}$ at $x = p$ passes through $(p, \sqrt[3]{p^2})$ , its equation comes to $y = \frac{2}{3\sqrt[3]{p}}x + \frac{1}{3}\sqrt[3]{p^2}$ .

To actually have three intersection points, line $d$ must be below the tangent line of $y = \sqrt[3]{x^2}$ at $x = p$ but above the point $(0, 0)$ . Therefore, for any value of $p$ (except $p = 0$ ), $m = \frac{2}{3\sqrt[3]{p}}$ and $0 < b < \frac{1}{3}\sqrt[3]{p^2}$ .

Substituting these values into $B$ gives $B < 3 - \frac{1}{(\frac{1}{3}\sqrt[3]{p^2})(\frac{2}{3\sqrt[3]{p}})^2}$ or $B < -\frac{15}{4}$ , which makes the inequality true .