Let $P(x) = (x-x_1)(x-x_2)...(x-x_n)$ where all the n roots are real and distinct. Let $y_1,y_2,...,y_{n-1}$ be the roots of $P'$. Show that
$\min_{i\neq j}|x_i-x_j|<\min_{i\neq j}|y_i-y_j|$.
My thoughts: We have $P'(x) = P(x)(\frac1{x-x_1}+...+\frac1{x-x_n})$. So we may consider $P'(x)/P(x)$, which has poles at the roots of $P$.
In a couple places below, I'll use the following fact without proof: if $f(x) = (x-x_1)(x-x_2)\cdots(x-x_n)$ where $x_1<x_2<\cdots<x_n$, then the roots $y_1\le y_2\le \cdots\le y_{n-1}$ of $f'$ are such that $ x_1<y_1<x_2<y_2<\cdots<y_{n-1}<x_n$.
Let $P(x) = (x-x_1)(x-x_2)\cdots(x-x_n)$. Assume without loss of generality $x_1<x_2<\cdots x_n$. Let $y_1<y_2<\cdots<y_{n-1}$ be the roots of $P'$, so $x_1<y_1<x_2<y_2<\cdots<y_{n-1}<x_n$.
Suppose $1\le m\le n-2$ and focus attention on $x_m < y_m < x_{m+1} < y_{m+1} < x_{m+2}$. I will show that $$ (1) \ \ \ 2(y_{m+1} - y_m) > x_{m+2} - x_m. $$ From this fact, it follows that either $x_{m+2} - x_{m+1}<y_{m+1}-y_m$ or $x_{m+1}-x_m<y_{m+1}-y_m$, since otherwise, there's no way that (1) could hold. This proves the result that $$ \min_{i\ne j} |x_i-x_j| < \min_{i\ne j} |y_i - y_j|, $$ since if $[y_m,y_{m+1}]$ is the shortest interval between $y_i$'s, then $[x_{m'},x_{m'+1}]$ is even shorter for $m'$ either $m$ or $m+1$.
Now (1) is a consequence of the following lemma:
Lemma: Suppose $a<b<c<d$ and $f:\bf{R} \to \bf{R}$ is strictly convex or strictly concave on $[a,d]$. Suppose further that $f(b) = f(c) = 0$ and $$ \int_a^d f(x) dx = 0. $$ Then $ 2(c - b) > d -a$.
Proof of Lemma: See the diagram below. Assume without loss of generality that $f$ is strictly convex. Since $\int_a^d f = 0$ and $f$ is strictly convex, $f(a) > 0$, $f(d)>0$, $f$ achieves a unique minimum at some $e\in (b,c)$, $f$ is strictly decreasing on $[a,e]$, and $f$ is strictly increasing on $[e,d]$. It also follows there is a $p\in (a,d)$ such that $$ \int_a^p f(x) dx = \int_p^d f(x) dx = 0. $$ Convexity implies there are lines $y=m_b(x-b)$ and $y=m_c(x-c)$ through $(b,0)$ and $(c,0)$ respectively such that $m_b < 0$, $m_c>0$, $$ (2)\ \ \ f(x) > m_b(x-b) $$ and $$ f(x) > m_c(x-c). $$ (See the diagram!)
Since $\int_a^p f(x) dx = 0$, (2) implies $$ (3) \ \ \ \int_a^p m_b(x-b) dx < 0. $$ Evaluating the integral (or an equivalent geometric argument that's obvious once one interprets inequality (3) with reference to the diagram) gives $$ p - b > b - a. $$ Similar considerations for $f$ on the interval $[p,d]$ demonstrates $$ c - p > d - c. $$ Adding the last two inequalities gives $$ c - b > (b - a) + (d - c). $$ Rearranging, we have $$ 2 (c - b) > d - a. $$ This proves the lemma.
Now consider the lemma with $f = P'$, $a = x_m$, $b = y_m$, $c=y_{m+1}$, and $d=x_{m+2}$. Note that $f=P'$ is indeed either strictly convex or strictly concave on $[x_m,x_{m+2}]$. To see this, observe that $P$ has exactly 3 roots in $[x_m,x_{m+2}]$, $P'$ has exactly 2, $P''$ has exactly 1, and $P'''$ has exactly 0. Thus, $P'''$ is either always positive or always negative on $[x_m,x_{m+2}]$. If $P'''>0$ on $[x_m,x_{m+2}]$, then $P'$ is strictly convex and it's strictly concave if $P'''<0$. Also note that $$ \int_{x_m}^{x_{m+2}} P'(x) dx = P(x_{m+2}) - P(x_m)=0. $$ Thus, the lemma applies and the inequality (1) above follows.
This completes the proof the desired result.