Show that the difference between the greatest and the least of them is not less than $\sqrt{a^2-3b}$ nor greater than $2\sqrt{a^2-3b}$ Let the roots of the cubic equation $x^3+ax^2+bx+c=0$ be real.Show that the difference between the greatest and the least of them is not less than $\sqrt{a^2-3b}$ nor greater than $2\sqrt{a^2-3b}$.

Let the real roots of the cubic equation $x^3+ax^2+bx+c=0$ be $\alpha,\beta,\gamma$.Let $\alpha\geq\beta\geq\gamma$.Then:
$$\alpha+\beta+\gamma=-a \tag1$$
$$\alpha\beta+\beta\gamma+\gamma\alpha=b \tag2$$
$$\alpha\beta\gamma=-c \tag3$$
I am stuck on how to prove
$$\sqrt{a^2-3b}\leq\alpha-\gamma\leq2\sqrt{a^2-3b}$$
Please help me out.Thanks.
 A: We have $a^2-2b=(\alpha+\beta+\gamma)^2-3(\alpha\beta+\beta\gamma+\gamma\alpha)=\alpha^2+\beta^2+\gamma^2-(\alpha\beta+\beta\gamma+\gamma\alpha)$. So we want to show that $\alpha^2+\beta^2+\gamma^2-(\alpha\beta+\beta\gamma+\gamma\alpha) \leq (\alpha-\beta)^2$. This is equivalent to $0\leq \alpha \beta+\beta \gamma -\alpha \gamma -\beta^2=(\beta-\alpha)(\gamma-\beta)$, which is true because $0\leq \alpha \leq \beta \leq \gamma$. If you square and expand the second inequality you see that it follows from $$3\alpha^2+4\beta^2+3\gamma^2-4\alpha\beta-4\beta\gamma-2\gamma\alpha\geq \alpha^2+2\beta^2+\gamma^2-2\alpha\beta-2\beta\gamma=(\alpha-\beta)^2+(\gamma-\beta)^2\geq 0$$
Here we used the well know fact that for real numbers $x,y,z$ we have $x^2+y^2+z^2\geq xy+yz+zx$
A: HINT
Suppose we let the minimum turning point be at $x_{2}$ and maximal turning point be at $x_{1}$. 
A quick sketch of the cubic function given reveals that:
$\alpha \geq x_{2} \geq \beta \geq x_{1} \geq \gamma$.
But since,
$f'(x)=3x^{2}+2ax+b$, we have,
$x_{1}=\frac{-a-\sqrt{a^{2}-3b}}{3}$ and $x_{2}=\frac{-a+\sqrt{a^{2}-3b}}{3}$.
Note that the distance along the x-axis between the turning points is at most the distance between the largest and smallest roots of the cubic. This makes sense if you look at the graph of this cubic.
Hence:
$x_{2}-x_{1} \leq \alpha - \gamma  \Rightarrow  \frac{-a+\sqrt{a^{2}-3b}}{3} -\frac{-a-\sqrt{a^{2}-3b}}{3} \leq \alpha - \gamma \Rightarrow  \frac{2\sqrt{a^{2}-3b}}{3} \leq \alpha - \gamma  \Rightarrow  \sqrt{a^{2}-3b} \leq \alpha - \gamma $ 
Now see what other properties of the turning point that you can use to find the upper bound for this distance.
