Link between polynomial and derivative of polynomial

Question

I can't seem to solve this problem, can anyone help me please? The problem is:

Let real numbers $a$,$b$ and $c$, with $a ≤ b ≤ c$ be the 3 roots of the polynomial $p(x)=x^3 + qx^2 + rx + s$. Show that if we divide the interval $[b, c]$ into six equal parts, then one of the root of $p'(x)$ (the derivative of the polynomial $p(x)$) will be in the 4th part.

What I did was:

Because we know the roots of $p(x)$ are $a$,$b$ and $c$, we can write the polynomial $p(x)$ like this: $p(x) = (x-a)(x-b)(x-c)$

So we have $p(x) = x^3 + qx^2 + rx + s = (x-a)(x-b)(x-c)$

We find the value of $q$,$r$ and $s$: $q = -(a+b+c)$

$r = (ab + ac + bc)$

$s = -abc$

We have that $p'(x) = 3x^2 + 2qx + r$ We want to find the roots of $p'(x)$, so if we apply the quadratic formula, we get: $(-2q ± 2*\sqrt{q^2 - 3r})/6$

Because we know the value of q,r and s, we can rewrite this expression like this: $(2(a+b+c) ± 2\sqrt{a^2 + b^2 + c^2 - ab - bc -ca})/6$

But I am stuck here, any help would be great. Thank you in advance.

Answer 1

Since $p(x) = (x-a)(x-b)(x-c)$,

$$\begin{align}p'(x) &= (x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)\\ &= (x-b)(x-c) + (x-a)(2x - (b+c))\end{align}$$ At $x = \frac{b+c}{2}$, we have

$$p'(x) = \frac{c-b}{2}\cdot\frac{b-c}{2} + (x-a)\cdot 0 = -\frac{(c-b)^2}{4} \le 0$$

At $x = \frac{b+2c}{3}$, we have

$$\begin{align}p'(x) &= \frac{2(c-b)}{3}\cdot\frac{b-c}{3} + \left(\frac{b+2c}{3} - a\right)\cdot\frac{c-b}{3}\\ &= -\frac{2(c-b)^2}{9} + \left(\frac{2(c-b)}{3} + (b-a)\right)\cdot\frac{c-b}{3}\\ &= \frac{(b-a)(c-b)}{3}\\&\ge 0\end{align}$$ This implies $p'(x)$ has a root in $\left[\frac{b+c}{2},\frac{b+2c}{3}\right]$.

Answer 2

I have an inelegant brute force solution. Since the problem is translation invariant, we may assume $a=0$ to simplify calculations so that $0\leq b\leq c$. It should be clear that the root of $p'(x)$ we are interested in is the largest of the two i.e. the one with the plus sign. We need to prove that $$\dfrac{3b+3c}{6}\leq\dfrac{2(b+c)+2\sqrt{b^2+c^2-bc}}{6}\leq\dfrac{2b+4c}{6}$$ On the left we obtain $$b+c\leq 2\sqrt{b^2+c^2-bc}$$ where both sides are non-negative, squaring gives $$b^2+c^2+2bc\leq 4(b^2+c^2-bc)\Rightarrow 3b^2+3c^2-6bc\geq 0$$ which holds. On the right we have $$2\sqrt{b^2+c^2-bc}\leq 2c$$ so that $$b^2+c^2-bc\leq c^2\Rightarrow b(b-c)\leq 0$$ which holds since $0\leq b\leq c$.