The Quartic Formula using Principal Roots: Cubic Theorem

Theorem. A cubic polynomial $y^3+By^2+Cy+D$ has three positive real roots if and only if $B<0$, $C>0$, $D<0$, and $(2B^3-9BC+27D)^2-4(B^2-3C)^3\leq0$.

Proof. If $y^3+By^2+Cy+D$ has three positive real roots then its discriminant $\Delta$ is nonnegative, so $-27\Delta=(2B^3-9BC+27D)^2-4(B^2-3C)^3\leq0$.

If the roots of the cubic are denoted $y_1$, $y_2$, and $y_3$, then using the factor theorem, expanding and equating coefficients yields: \begin{align*} B &= -(y_1 + y_2 + y_3) \\ C &= y_1y_2 + y_1y_3 + y_2y_3 \\ D &= - y_1y_2y_3 \end{align*} From which it follows that $B<0$, $C>0$, and $D<0$, since $y_1$, $y_2$, $y_3>0$.

Conversely, if $B<0$, $C>0$, $D<0$, and $(2B^3-9BC+27D)^2-4(B^2-3C)^3\leq0$, then the discriminant of $y^3+By^2+Cy+D$ is nonnegative, so the cubic has three real roots.

Suppose $y\leq0$; then $y^3\leq0$, $By^2\leq0$, $Cy\leq0$, and $D<0$, so \[ y^3 + By^2 + Cy + D < 0 . \] Thus $y^3+By^2+Cy+D$ has no nonpositive roots, so its three real roots must be positive.$\Box$