Is It Possible to Prove Real Root Properties in Cubic Equations?

[anemone]

Here is this week's POTW:

-----

The cubic equation $x^3+ax^2+bx+c=0$ has real coefficients and three real roots. Show that $a^2-3b\ge 0$ and that $\sqrt{a^2-3b}$ is less than or equal to the difference between the largest and the smallest roots.

-----

 

[anemone]

No one answered last week's POTW, (Sadface) but you can read the suggested solution as follows:

Spoiler: Suggested solution of other

Let the three real roots be $r,\,s$ and $t$ such that $r\ge s \ge t$.

$a^2-3b=(r+s+t)^2-3(rs+st+tr)=r^2+s^2+t^2-rs-st-tr=\dfrac{(r-s)^2}{2}+\dfrac{(s-t)^2}{2}+\dfrac{(t-r)^2}{2}\ge 0$, with equality when $r=s=t$.

Also, $r\ge s \ge t$ implies

$(s-r)(s-t)\le 0\\ s^2-rs-st+rt \le 0 \\ (r^2+s^2+t^2)-(rs+st+tr) \le r^2+t^2-2rt \\ a^2-3b \le (r-t)^2 \\ \sqrt{a^2-3b} \le r-t $