Cubic equation problem $\frac{x^3}{3}-x=k$

Question

The cubic function $\frac{{{x^3}}}{3} - x = k$ has three different roots $\alpha,\beta,\gamma$ about the real number k. Let's call the minimum value of $|\alpha|+|\beta|+|\gamma|$ as $m$. FInd the value of $m^2$.

My approach is as follow

$\frac{{{x^3}}}{3} - x = k \Rightarrow f\left( x \right) = \frac{{{x^3}}}{3} - x - k$

$f'\left( x \right) = {x^2} - 1 = 0$

Hence $x = \pm 1$

$f\left( 1 \right) = \frac{1}{3} - 1 - k = - \left( {k + \frac{2}{3}} \right)\& f\left( { - 1} \right) = - \frac{1}{3} + 1 - k = - \left( {k - \frac{2}{3}} \right)$

For real roots $f\left( 1 \right)f\left( { - 1} \right) < 0$

Therefore $ - \left( {k + \frac{2}{3}} \right) \times \left( { - \left( {k - \frac{2}{3}} \right)} \right) < 0$

$k \in \left( { - \frac{2}{3},\frac{2}{3}} \right)$

We know that $\alpha+\beta+\gamma=0$

But how we will find the minimum values of the sum of the modulus of the roots.

Answer 1

Hint $\;-\;$ using $\,\alpha\beta+\beta\gamma+\gamma\alpha=-3\,$:

$$ \require{cancel} \begin{align} m^2 &= \left(|\alpha|+|\beta|+|\gamma|\right)^2 \\ &= \alpha^2+\beta^2+\gamma^2 + 2\left(|\alpha\beta|+|\beta\gamma|+|\gamma\alpha|\right) \\ &= \cancel{\left(\alpha+\beta+\gamma\right)^2} - 2(\alpha\beta+\beta\gamma+\gamma\alpha)+ 2\left(|\alpha\beta|+|\beta\gamma|+|\gamma\alpha|\right) \\ &= 6 + 2\left(|\alpha\beta|+|\beta\gamma|+|\gamma\alpha|\right) \\ &\ge 6 + 2\,|\alpha\beta+\beta\gamma+\gamma\alpha| \\ &= 12 \end{align} $$

This gives a lower bound on $m^2\,$. To prove it's an actual minimum, it is enough to find a $k$ such that $|\alpha|+|\beta|+|\gamma| = 2 \sqrt{3}\,$, which turns out not to be too hard.

Note: the above assumes the roots are real (per OP's comments), in order for $\,|\alpha|^2=\alpha^2\,$ to hold.

Answer 2

Note: This is assuming that the question intends to ask that each of $\{\alpha,\beta,\gamma\}$ must be real in addition to $k$. If they are allowed to be non-real, the situation is trickier.

The sum $|\alpha|+|\beta|+|\gamma|$ is preserved if $k$ is replaced by $-k$, so we can without loss of generality assume that $k<0$, so that there are two positive roots and one negative root. Let the positive roots be $\alpha>\beta$, and let $\alpha(k)$ and $\beta(k)$ represent the functions mapping $k$ to the largest root and to the second largest root of $x^3/3-x-k$, respectively. We seek to minimize $2\alpha(k)+2\beta(k)$.

See if you can do this via calculus, writing the derivative of $2\alpha(k)+2\beta(k)$ in terms of $\alpha(k)$, $\beta(k)$, and $k$. Then you only need to check the endpoints ($k=0$ and $k=-2/3$), and any places where the derivative may be $0$.

Answer 3

Note: this is similar to another answer, except that by recognizing only one root needs to be tracked instead of the other two the calculation is simplified.

First observe that if we reverse the sign of $k$ then all roots are also reversed in sign with no effect on the set of absolute values, thus no impact on the sum $|\alpha|+|\beta|+|\gamma|$. So we can cover all values of this sum by considering just the case where $k$ is nonpositive.

Then only one root is negative by Descartes' Rule of Signs. Calling that root $\alpha$ we then have

$|\alpha|+|\beta|+|\gamma|=-\alpha+\beta+\gamma$

$\color{blue}{=-2\alpha\because\alpha+\beta+\gamma=0}$

Thus find the nonpositive value of $k$ that minimizes the absolute value of the negative root $\alpha$. Given that $k=0$ gives $\alpha=-\sqrt3$ and $k<0$ must give $\alpha<-\sqrt3$ (why?), you then get your answer using the blue equation above.