If all convex combinations of $p(x)$ and $q(x)$ have real roots, then $p,q$ have a common interlacing poly I heard this result in a talk the other day:
Suppose $p$ and $q$ are polynomials. Suppose $p$ is a polynomial of degree $n$ and $q$ a polynomial of degree $n-1$. Call $q$ an interlacer of $p$ if the roots $a_i$ of $p$ and $b_i$ of $q$ are such that
$$a_1 \leq b_1 \leq a_2 \leq b_2 \leq \dotsb \leq b_{n-1} \leq a_n.$$
Suppose $a,b$ are polynomials of the same degree such that $\lambda a(x) + (1-\lambda )b(x)$ has only real roots for all $\lambda \in [0,1]$. Then $a$ and $b$ have a common interlacing polynomial.
I've been thinking about how to prove this. Does anyone have an idea?
 A: As Stephen Montgomery-Smith says in the comments, this is Lemma 3.5 in Marcus-Spielman-Srivastava. They credit it to several different sources, of which Dedieu (Theorem 2.1) is the earliest and Fell (Theorem 2') is to my mind the clearest. Fell also has the advantage of being freely available online.
Here is another presentation of Fell's argument, which handwaves a few "geometrically obvious" details, but seems pretty to me.
For simplicity, assume that $a$ and $b$ have no roots in common. Set $h(x) = \frac{b(x)}{b(x)-a(x)}$. So $x$ is a root of $\lambda a(x) + (1-\lambda) b(x) =0$ if and only if $\lambda = h(x)$. Let $\Gamma$ be the graph of $h$ in the horizontal strip $\mathbb{R} \times [0,1]$. By hypothesis, for every $\lambda \in [0,1]$, the graph $\Gamma$ crosses the line $\mathbb{R} \times \{ \lambda \}$ at $n$ points. Let those $n$ points be $r_1(\lambda)$, $r_2(\lambda)$, ..., $r_n(\lambda)$. Let $I_k = \{ r_k(\lambda) : \lambda \in [0,1] \}$. In other words, $I_k$ is the interval beneath the arc of $\Gamma$ from $(0,r_k(0))$ to $(1, r_k(1))$. 
We claim that the intervals $I_1$, $I_2$, ..., $I_n$ are disjoint. Suppose to the contrary that $x \in I_i \cap I_j$. Then the vertical line $\{ x \} \times \mathbb{R}$ meets $\Gamma$ twice, so $\Gamma$ violates the vertical line test. 
So the closed intervals $I_j$ are disjoint. Choose $n-1$ points $s_1$, $s_2$, ..., $s_{n-1}$ with $s_j$ separating $I_j$ and $I_{j+1}$. Then the polynomial $\prod (x-s_j)$ interlaces all the polynomials $\lambda a(x) + (1-\lambda) b(x)$.
