I'm taking capital $A=1$ and then $a < b < c$ as the roots. we get the cubic
$$ x^3 - \sigma_1 x^2 + \sigma_2 x - \sigma_3 \; , $$
where
$$ \sigma_1 = a + b + c, $$
$$ \sigma_2 = bc + ca + ab, $$
$$ \sigma_3 = abc. $$
The first derivative is $3 x^2 - 2 \sigma_1 x + \sigma_2,$ with roots
$$ \frac{\sigma_1 \pm \sqrt{\sigma_1^2 - 3 \sigma_2}}{3} \; .$$
Worth emphasizing that, with $a,b,c$ distinct, we get
$$ \sigma_1^2 - 3 \sigma_2 = a^2 + b^2 + c^2 - bc - ca - ab = \frac{1}{2} \left( (b-c)^2 + (c-a)^2 + (a-b)^2 \right) $$
being strictly positive, so the two roots of $3 x^2 - 2 \sigma_1 x + \sigma_2$ are real and distinct.
The claim that $c > \frac{\sigma_1 + \sqrt{\sigma_1^2 - 3 \sigma_2}}{3}$ comes to combining the observation that $c > \frac{\sigma_1}{3}$ along with
$$ \left( c - \frac{\sigma_1}{3} \right)^2 -\left( \frac{\sigma_1^2 - 3 \sigma_2}{9} \right) = \frac{1}{3} (c-a)(c-b) > 0 $$
The claim that $a < \frac{\sigma_1 - \sqrt{\sigma_1^2 - 3 \sigma_2}}{3}$ comes to combining the observation that $a < \frac{\sigma_1}{3}$ along with
$$ \left( a - \frac{\sigma_1}{3} \right)^2 -\left( \frac{\sigma_1^2 - 3 \sigma_2}{9} \right) = \frac{1}{3} (c-a)(b-a) > 0 $$
The final claim is that $b$ lies between the critical points, in that the distance between $b$ and $\sigma_1/3$ is smaller than $\sqrt{\sigma_1^2 - 3 \sigma_2}/3.$ Indeed,
$$ \left( b - \frac{\sigma_1}{3} \right)^2 -\left( \frac{\sigma_1^2 - 3 \sigma_2}{9} \right) = \frac{1}{3} (c-b)(a-b) < 0 $$
Best Answer
Suppose that (including multiplicity) the roots of $$f(x) = A x^3 + B x^2 + C x + D,$$ $A \neq 0$, are $r_1, r_2, r_3$. Consider the quantity $$\Delta(f) := A^4 (r_3 - r_2)^2 (r_1 - r_3)^2 (r_2 - r_1)^2,$$ called the (polynomial) discriminant of $f$.
If $r_1, r_2, r_3$ are all real and pairwise distinct, then we see that $\Delta(f) > 0$. On the other hand, if $f$ has a repeated root, then $\Delta(f) = 0$, and if $f$ has one real root and two nonreal (necessarily conjugate) roots, substituting gives $\Delta(f) < 0$. (The coefficient $A^4$ is unnecessary for $\Delta$ to enjoy these properties, but among other things, its inclusion makes the below formula nicer.)
These three cases exhaust all of the possibilities, so we conclude:
$$\color{#bf0000}{\fbox{A cubic polynomial $f$ has three distinct, real roots iff $\Delta(f) > 0$.}}$$
The above definition of $\Delta$ is not immediately practical, since explicit formulas for the roots of a general cubic polynomial are unwieldy. On the other hand, with some work (say, by expanding and using Newton's Identities and Vieta's Formulas) we can write $\Delta(f)$ as a homogeneous quartic expression in the coefficients $A, B, C, D$: $$\Delta(f) = -27 A^2 D^2 + 18 ABCD - 4 A C^3 - 4 B^3 D + B^2 C^2.$$ This formula gives a computationally practical answer to the question:
$$ \bbox[0.5mm,border: 1px solid #bf0000]{ \color{#bf0000}{ \begin{array}{c} \textrm{A cubic polynomial} \\ f(x) = A x^3 + B x^2 + C x + D \\ \textrm{has three distinct, real roots iff} \\ -27 A^2 D^2 + 18 ABCD - 4 A C^3 - 4 B^3 D + B^2 C^2 > 0 \textrm{.} \end{array} } } $$
It's apparent that one can generalize the notion of discriminant to polynomials $p$ of any degree $> 1$, producing an expression homogeneous of degree $2(\deg p - 1)$ in the polynomial coefficients. In each case, up to a constant that depends on the degree and the leading coefficient of $f$, $\Delta(f)$ is equal to the resultant $R(f, f')$ of $f$ and its derivative.
By making a suitable real, affine change of variables $x \rightsquigarrow y = a x + b$, by the way, one can transform any given real cubic polynomial to the so-called depressed form $$\tilde{f}(y) = y^3 + P y + Q .$$ such transformations do not change the number of real roots or the multiplicities of roots. For a cubic polynomial in this reduced form the discriminant takes the simpler and well-known form $$\Delta(\tilde f) = -4 P^3 - 27 Q^2.$$