A curious corollary: there always exists some $c$ such that $2g'(c)h'(c)+g(c)h''(c)=0$ In a recent question, it was proved that if $f,g:[a,b]\to\mathbb R$ are two twice differentiable functions such that $f(a)=f(b)=g(a)=g(b)=0$ and both $g$ and $g''$ are non-vanishing on $(a,b)$, then $\frac{f(c)}{g(c)}=\frac{f''(c)}{g''(c)}$ for some $c\in(a,b)$.
Now, given any $g$ that satisfies the above premises, if we take $f=gh$, where $h$ is some twice differentiable function, then $f$ is also twice differentiable and the condition that $f(a)=f(b)=0$ is automatically satisfied. Therefore, by the above result,
\begin{aligned}
\frac{f(c)}{g(c)}=\frac{f''(c)}{g''(c)}
&\ \Rightarrow\ h(c)=\frac{g''(c)h(c)+2g'(c)h'(c)+g(c)h''(c)}{g''(c)}\\
&\ \Rightarrow\ g''(c)h(c)=g''(c)h(c)+2g'(c)h'(c)+g(c)h''(c)\\
&\ \Rightarrow\ 2g'(c)h'(c)+g(c)h''(c)=0.\\
\end{aligned}
Hence we obtain the following corollary:

Let $g:[a,b]\to\mathbb R$ be a twice differentiable function such that $g(a)=g(b)=0$ and both $g$ and $g''$ are non-vanishing on $(a,b)$. For every twice differentiable function $h:[a,b]\to\mathbb R$, there exists a point $c\in(a,b)$ such that
$$ \bbox[5px,border:2px solid red]
{
2g'(c)h'(c)+g(c)h''(c)=0.
}
$$

I find this corollary more intriguing than the original problem, because $h$ is basically unrelated to $g$. Are there any direct or intuitive proofs that do not refer to the original problem? Can the assumptions that $g$ and $g''$ are non-vanishing be weakened or even removed from the corollary? (In particular, since the boxed statement does not involve $g''$, I wonder if the corollary is still true when $g$ is only differentiable or $C^1$.)
 A: That corollary can be proved directly by applying Rolle's theorem to $g^2h'$. That is not surprising because the statement about $\frac{f(c)}{g(c)}=\frac{f''(c)}{g''(c)}$ is proved by applying Rolle's theorem to $\phi = f'g - fg'$. Substituting  $f=gh$ gives $\phi =g^2h'$.
The non-vanishing of $g''$  on $(a, b)$ in the other question guarantees that one can divide by $g''(c)$, so that condition is not needed here.
Since the corollary only involves $h'$ and the second derivative of $g$ is not needed, it would be more naturally stated as

Let $g, h:[a,b]\to\mathbb R$ be differentiable functions such that $g(a)=g(b)=0$ and $g$ is non-vanishing on $(a,b)$. Then there exists a point $c\in(a,b)$ such that
$$ 
2g'(c)h(c)+g(c)h'(c)=0.
$$

By applying Rolle's theorem to $g^\alpha h$ with $\alpha > 0$ it can be generalized as follows:

Let $g, h:[a,b]\to\mathbb R$ be differentiable functions such that $g(a)=g(b)=0$ and $g$ is non-vanishing on $(a,b)$. For every $\alpha > 0$ there exists a point $c\in(a,b)$ such that
$$ 
\alpha g'(c)h(c)+g(c)h'(c)=0.
$$

