Cubic inequality $x^3-bx^2+1>0$

Question

$$x^3-bx^2+1>0,$$

in which $b \in \mathbb{R}$. Is there a way to solve this? Thanks in advance.

Answer 1

This is just a partial answer. Assume you know the solutions of the cubic equation $$x^3-bx^2+1=0.$$

We will label the solutions as $x_1$, $x_2$ and $x_3$.

Then by the theorem of Vieta, we know that the left-hand side of the inequality can be written as

$$(x-x_1)(x-x_2)(x-x_3)>0.$$

Each term in the product can have positive and negative signs. The product of three numbers is positive if all individual terms are positive or if exactly two are negative.

In the case of complex conjugate roots, the left-hand side can be written as

$$(x-x_1)(x^2+\alpha x+\beta)>0.$$

In this case, the quadratic always has the same sign (it should be positive), because the quadratic doesn't have real roots. Hence, $(x-x_1)>0$ must be valid.

The problem is that the solutions of the cubic are not trivial expressions in $b$.

Answer 2

It is helpful to look at the extrema. You have $$ f(x) = x^3-bx^2+1 $$ and extrema at $$ 0 = f'(x) = x(3x- 2b)$$ so extrema are at $x=0$ and $x=2b/3$ with values $f(0) = 1$ and $f(2b/3) = 1- 4/27 b^3$. Hence you know that both extrema have values $> 0$ iff $b < 3/\sqrt[3]{4} \simeq 1.89$ and in this case, $f(x) \ge 0$ will hold for all $x > x_n$ where $x_n <0$ is the only root of $f(x)$.

If $b = 3/\sqrt[3]{4} \simeq 1.89$ there will be 2 real roots and you have the same solution as above, with additionally $f(x) = 0$ at $x = 2b/3 = 2/\sqrt[3]{4} \simeq 1.26$.

Conversely, iff $b > 3/\sqrt[3]{4} \simeq 1.89$ there will be 3 real roots and $f(x) \ge 0$ will be split: it will hold for two intervals: for $x > q$ where $q > 2b/3 > 2/\sqrt[3]{4} \simeq 1.26$, and for $x_n < x < r$ where $x_n <0$ and $ 0 <r < 2b/3$. Here, $x_n, q, r$ are the three roots of $f(x)$.