for $n\ge 2$ find the largest real number $C(n)$ so that for all $f\in \mathcal{P}_n, d(f') \ge C(n) d(f)$ 
Let $\mathcal{P}_n$ denote the set of all polynomials of degree n with all real roots. For $f\in \mathcal{P}_n$ for some $n\ge 1,$ let $d(f)$ be the distance between its largest real root and its smallest real root. For any $n\ge 2,$ find the largest real number $C(n)$ so that for all $f\in \mathcal{P}_n, d(f') \ge C(n) d(f)$.

It's known that for a nonconstant polynomial $f, \dfrac{f(x)}{f'(x)}=\sum_{i=1}^k \dfrac{1}{x-r_i}$ where $r_1 < r_2<\cdots < r_k$ are the roots of $f$. A hint for this problem was that one can obtain a polynomial $g$ with $d(g')\leq d(f')$ by moving two consecutive roots closer to each other (maybe if $r_1 < r_2$ are consecutive roots of $f$ we can achieve the result by making $(r_1+r_2)/2$ a double root of $g(x)$ or we can choose two roots $r_1'$ and $r_2'$ with $r_1 < r_1' <(r_1+r_2)/2 < r_2' < r_2$). But which two consecutive roots should be moved closer to each other? Some tricks I've learned when dealing with polynomials include unique factorization (not unlike the same concept applied to integers), Vieta's formulas, finding the greatest common divisor of two polynomials, using inequalities, using Rolle's theorem, using the fact that the roots of a polynomial are in fact a continuous function of the coefficients (which seems nontrivial to prove), using the fact that the multiplicity of a root in $f(x)$ is one larger than its corresponding multiplicity in $f'(x)$, etc. For the case when $n=2$, the polynomials in $\mathcal{P}_n$ are just the quadratic polynomials and $d(f') = 0$ for every quadratic polynomial f. Note that every polynomial with real coefficients has an even number of nonreal roots. Now let $f(x) = ax^3+bx^2+cx+d, a\neq 0$ have all real roots. Write $f(x)=a(x-r_1)(x-r_2)(x-r_3)$ where $r_1\leq r_2\leq r_3$ are its roots. $d(f) = r_3 - r_1$. $f'(x)= a[(x-r_1)+(x-r_2)]$ and $d(f') = 0.$ It might simplify things for this problem if one narrows down the search for polynomials in $\mathcal{P}_n$ by only considering some polynomials (e.g. polynomials where all the roots are distinct).
 A: We will find the minimal possible $d(f')/d(f)$ under the assumption that $f$ has the degree $n\geq 3$ and has all real roots.
This is a problem regarding the distance of roots, so we may normalize the leading coefficient of $f$ by $1$. Also, we assume that $d(f)=2$ with the largest root $1$ and the smallest root $-1$.
Then we now focus on finding the optimal choice of $f$ such that $d(f')/d(f)$ is minimized.  One observation is that if one of the roots $\pm 1$ are simple (exponent $1$'s on $x+1$ or $x-1$), then we have $d(f')<d(f)$. If both roots $\pm 1$ are not simple, then we have $d(f')=d(f)$. Thus, without loss of generality, assume that $-1$ is simple.
Let $f(x)=(x+1)(x-r_1)^{e_1}\cdots (x-r_{k-2})^{e_{k-2}}(x-1)^{e_{k-1}}$ with $e_i\ge1$ and $r_i$'s are distinct with $r_1<\cdots <r_{k-2}$, and $1+e_1+\cdots+e_{k-1}=n$.
Let $s_1$ be the smallest real root of $f'(x)=0$. Then necessarily $-1<s_1<r_1$. Since
$$
\frac{f'(x)}{f(x)}=\frac1{x+1} + \frac{e_1}{x-r_1}+ \cdots + \frac{e_{k-2}}{x-r_{k-2}} + \frac{e_{k-1}}{x-1},$$
we have
$$
\frac1{s_1+1}=\frac{e_1}{r_1-s_1}+\cdots + \frac{e_{k-2}}{r_{k-2}-s_1}+\frac{e_{k-1}}{1-s_1}. $$
and thus
$$
1=\frac{e_1(s_1+1)}{r_1-s_1}+\cdots + \frac{e_{k-2}(s_1+1)}{r_{k-2}-s_1}+\frac{e_{k-1}(s_1+1)}{1-s_1}. \ \ \ \ (*)
$$
We move $r_1$ toward $r_2$ while fixing $s_1$ and $r_2, \ldots, r_{k-2}$. This makes the RHS of $(*)$ smaller. Note that
$$
\frac{e(s+1)}{r-s}=-e+\frac{e(r+1)}{r-s}
$$
is an increasing function of $s$ when $s<r$. Applying this to all terms of the RHS of $(*)$, $s_1$ must be made larger to match the LHS $1$ if we move $r_1$ toward $r_2$.
Thus, merging the roots $r_1$ and $r_2$ with the root $r_2$ with multiplicity $e_1+e_2$ is a better choice than keeping them separate. To keep enlarging $s_1$, we will need to merge all roots to $1$, giving $f(x)=(x+1)(x-1)^{n-1}$. Then $f'(x)=(x-1)^{n-2}(nx+n-2)$ and $d(f')/d(f)=(n-1)/n$.
This is the minimal $d(f')/d(f)$ given that the root $1$ of $f$ is not simple, i. e., $e_{k-1}>1$.
Suppose now the root $1$ of $f$ is also simple, i. e. $e_{k-1}=1$. Let $s_{k-1}$ be the largest root of $f'(x)=0$. Then necessarily $r_{k-2}<s_{k-1}<1$. Similarly,
$$
1=\frac{1-s_{k-1}}{s_{k-1}+1}+\frac{e_1(1-s_{k-1})}{s_{k-1}-r_1}+\cdots+\frac{e_{k-2}(1-s_{k-1})}{s_{k-1}-r_{k-2}}, \ \ \ (**)
$$
As in the given hint of the problem, we look for merging the consecutive roots of $f(x)=0$. Consider the roots $r_{i}$, $r_{i+1}$ between $-1$ and $1$. The terms of $(*)$ corresponding to these roots are
$$
\frac{e_i(s_1+1)}{r_i-s_1}+\frac{e_{i+1}(s_1+1)}{r_{i+1}-s_1}  \ \ \ \rm (I)
$$
The terms of $(**)$ corresponding to these roots are
$$
\frac{e_i(1-s_{k-1})}{s_{k-1}-r_i}+\frac{e_{i+1}(1-s_{k-1})}{s_{k-1}-r_{i+1}} \ \ \ \rm (II)
$$
We look for $r_*$ to merge $r_i$ and $r_{i+1}$ to a root $r_*$ with multiplicity $e_i+e_{i+1}$. We have $r_{I*}$ with $r_i<r_{I*}<r_{i+1}$ such that
$$
\frac{e_i }{r_i-s_1}+\frac{e_{i+1} }{r_{i+1}-s_1}=\frac{e_i+e_{i+1}}{r_{I*}-s_1}.
$$
Then $s_1$ must be taken smaller if $r_i<r<r_{I*}$, and $s_1$ must be taken larger if $r_{I*}<r<r_{i+1}$.
Similarly, there is $r_{II*}$ with $r_i<r_{II*}<r_{i+1}$ such that
$$
\frac{e_i }{s_{k-1}-r_i}+\frac{e_{i+1} }{s_{k-1}-r_{i+1}}=\frac{e_i+e_{i+1} }{s_{k-1}-r_{II*}}.
$$
Then $s_{k-1}$ must be taken smaller if $r_i<r<r_{II*}$, and $s_{k-1}$ must be taken larger if $r_{II*}<r<r_{i+1}$.
Since $1/(x-s_1)$ is decreasing concave up for $s_1<x<s_{k-1}$ and $1/(s_{k-1}-x)$ is increasing concave up for $s_1<x<s_{k-1}$, we must have $r_{I*}<r_{II*}$. See the attached picture for an explanation.

Then any choice of $r_*$ within $r_{I*}\leq r_* \leq r_{II*}$ makes
the resulting polynomial's $d(f')$ at most the $d(f')$ from the original polynomial.
Repeating this procedure of merging the consecutive roots, the optimal choice is made after merging all roots between $-1$ and $1$. Then we must have
$$
f(x)=(x^2-1)(x-r)^{n-2}
$$
with $-1<r<1$.
This gives
$$
f'(x)=(x-r)^{n-3}(nx^2-2rx-(n-2)).
$$
The quadratic factor has roots
$$
x=\frac{r\pm \sqrt{r^2+n(n-2)}}n.
$$
Then we have $d(f')=\frac{2\sqrt{r^2+n(n-2)}}n$ and
$$
\frac{d(f')}{d(f)}=\frac{\sqrt{r^2+n(n-2)}}n.
$$
Now this expression is minimal if $r=0$. That means the optimal choice is
$$
f(x)=(x^2-1)x^{n-2}
$$
and we have
$$
\frac{d(f')}{d(f)}=\sqrt{\frac{n-2}n} < \frac{n-1}n.
$$
Therefore, the largest $C(n)$ is the minimal $d(f')/d(f)$ in all cases, is $\sqrt{\frac{n-2}n}$.
