Link between polynomial and derivative of polynomial

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 $\left[\frac{b+c}{2},\frac{b+2c}{3}\right]$.

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 $0\leq b\leq c$.