Inequality based on cubic equation

Question

If $2x^3+ax^2+bx+4=0$ (where $a$ and $b$ are positive and real) has three real roots,
then prove that $a \geq 6\sqrt[3]{2}$.

How to approach this question?

Answer 1

This is a consequence of Newton's or MacLaurin's inequalities. If $\zeta_1,\zeta_2,\zeta_3$ are the real roots of the given cubic polynomial, by Vieta's formulas $$\zeta_1+\zeta_2+\zeta_3 = -\frac{a}{2}\qquad\zeta_1\zeta_2+\zeta_1\zeta_3+\zeta_2\zeta_3 = \frac{b}{2}\qquad \zeta_1 \zeta_2\zeta_3=-2$$ and by Descartes' rule of signs $\zeta_i<0$. So the polynomial $2x^3-ax^2+bx-4$ has three positive real roots and its elementary symmetric means are $$ S_1 = \frac{a}{6},\qquad S_2=\frac{b}{6},\qquad S_3=2 $$ where $S_1\geq \sqrt[3]{S_3}$ must hold, leading to $\color{red}{a\geq 6\sqrt[3]{2}}$ as wanted.