I am having trouble proving the Obreschkoff-Hermite-Kakeya Theorem:
Theorem (OBK). Let $P$ and $Q$ be two non-constant real valued polynomials with no common zeros. $P$ and $Q$ have only real zeros that strictly interlace, if and only if, $\lambda P(z)+\mu Q(z)$ has only real simple zeros for every $\lambda,\mu\in\mathbb{R}$. Strictly interlacing means that between every two zeros of $P$ is a zero of $Q$ and visa-versa, between every two zeros of $Q$ is a zero of $P$. Strictly interlacing immediately implies that $|\deg(P)-\deg(Q)|\le 1$. To avoid trivialities we take the $0$ function to have simple real zeros.
The Wronskian Theorem of Interlacing (don't know what else to call it) seems useful:
Theorem (WRO). Let $P$ and $Q$ be two real valued polynomials with only real zeros. $P$ and $Q$ have strictly interlacing zeros, if and only if, $$W[f(x),g(x)]=f'(x)g(x)-g'(x)f(x)\not=0$$ for every $x\in\mathbb{R}$. Notice, that the Wronskian is continuous so we really have $W[f,g]> 0$ on the whole real line or $W[f,g]<0$ on the whole real line.
I originally thought that OBK followed from WRO, but I am having trouble completing the proof.
(OBK$\Rightarrow$) WLOG, suppose that $\deg(P)\le\deg(Q)$. If $P$ and $Q$ have only real strictly interlacing zeros then $W[P,Q]\not=0$. However, a calculation shows that $\lambda \cdot W[P,Q]=W[\lambda P+\mu Q,Q]$, thus $\lambda P + \mu Q$ strictly interlaces with $Q$. So since $\deg(\lambda P + \mu Q)=\deg(Q)$, $\lambda P + \mu Q$ has only real simple zeros.
(OBK$\Leftarrow$) This is the direction I can't seem to complete. Assuming $\lambda P + \mu Q$ has only real simple zeros for every $\lambda,\mu$, how do I show that $P$ and $Q$ have interlacing zeros? Showing that $|\deg(P)-\deg(Q)|\le 1$ even seems challenging. Although I can accomplish that at least. Supposing $\deg(P) < \deg(Q)$, we consider the graph of $\frac{P}{Q}$ which has a horizontal asymptote at zero, visually we must note that the approach to the asymptote can occur from under the x-axis or above the x-axis. Since the zeros of $P$ are simple, then for small $\epsilon$ the horizontal lines $y=\pm\epsilon$ will cross the graph $\frac{P}{Q}$ at least $\deg(P)$ times where the zeros of $P$ occur. However, $y=\pm\epsilon$ could cross the graph more that $\deg(P)$ times depending on the approach of $\frac{P}{Q}$ to it's horizontal asymptote (the x-axis). But by assumption $P+\epsilon Q$ and $P-\epsilon Q$ both have $\deg(Q)$ real simple zeros, that is $\frac{P}{Q}+\epsilon$ and $\frac{P}{Q}-\epsilon$ have $\deg(Q)$ zeros, which is the same number of zeros. This forces both $y=\epsilon$ and $y=-\epsilon$ to cross the graph of $\frac{P}{Q}$ exactly $\deg(P)+1$ times. So we have $\deg(P)+1=\deg(Q)$.
So we now know $|\deg(P)-\deg(Q)|\le 1$, but this is as far as I have gotten. I still can't seem to show that they have interlacing zeros. In addition the above argument I just gave, is terribly unsophisticated and practically requires some pictures to be drawn.
Any insight would be greatly appreciated. A reference to this theorem would be great as well.
Theorem 6.3.8 of Analytic Theory of Polynomials has a proof.
For a more elementary proof, you can look at Proposition 1.35 of Polynomials, roots and interlacing