Is there any hope to prove that $g(x)>-4$ if $f(x)<0$? I have these two functions for $x>0$, $\beta>0$ and $\alpha$ (all reals)
$$ f(x)= \frac{\alpha \; \sin (\beta \; x)}x+4    \cos (\beta\;  x) ,\qquad\qquad\qquad\qquad\qquad\qquad\\
g(x)=\frac{\alpha  \cos (\beta  \;x)}{\beta \; x^2}+\frac{\alpha  \cos (2 \beta \; x)}{\beta \; x^2}-\frac{\alpha\; \cot  (\frac{\beta \; x}{2})}x    -\frac{4\;\sin (2 \beta \; x)}{\beta\;x}   
$$

Numerically, I am sure that if $f(x)<0$, then $g(x)>-4$ (a plot of a typical example is attached); I appreciate any hints (if there are any) to prove this analytically.


 A: $\newcommand{\R}{\mathbb R}$Your conjecture is true.
Indeed,
\begin{equation*}
\begin{aligned}
    f(x)&=F(a,u):=a\frac{\sin u}u+4\cos u, \\ 
    g(x)&=G(a,u):=a\frac{\cot(u/2)}{u^2}\,h(u)-4\frac{\sin 2u}u, 
\end{aligned}
\end{equation*}
where
\begin{equation*}
    a:=\alpha\beta\in\R,\quad u:=\beta x>0, 
\end{equation*}
\begin{equation*}
    h(u):=\sin2u-\sin u-u. 
\end{equation*}
(One of the ways to check the equality $g(x)=G(a,u)$ is by using the identities $\sin u=\frac{2t}{1+t^2}$ and $\cos u=\frac{1-t^2}{1+t^2}$, where $t:=\tan\frac u2$, as well as the identities $\cos2u=2\cos^2u-1$ and $\sin2u=2\sin u\cos u$.)
So, your conjecture can be rewritten as follows:
for $a$ and $u$ as specified above,
\begin{equation*}
    F(a,u)<0\implies G(a,u)>-4. \tag{1}\label{1}
\end{equation*}
Note that $F(a,u)=4\not<0$ if $u=2k\pi$ for a natural $k$. So, without loss of generality (wlog) $u\ne 2k\pi$ for any natural $k$, and hence $\cot(u/2)\ne0$ and, moreover,
\begin{equation*}
    \cot(u/2)=\frac{\sin u}{2\sin^2(u/2)}\overset{\text{sign}}=\sin u, \tag{2}\label{2}
\end{equation*}
where $\overset{\text{sign}}=$ means the equality in sign.
Also, if $\sin u=0$, then $G(a,u)=0>-4$, so that implication \eqref{1} holds. So, wlog $\sin u\ne0$ and hence $\cos u\ne1$.
Lemma 1: $h<0$.
Lemma 2: $r(u):=\dfrac{u-\sin u}u<1.25$.
These elementary lemmas will be proved later in this answer.
Letting $a_u:=-4u\cot u$, note that

*

*$F(a,u)<0\iff a<a_u$ $\quad\text{if}\quad$ $\sin u>0$;


*$F(a,u)<0\iff a>a_u$ $\quad\text{if}\quad$ $\sin u<0$.
So, by Lemma 1 and \eqref{2},
\begin{equation*}
    F(a,u)<0\implies G(a,u)>G(a_u,u)=H(u):=R(u) r(u), 
\end{equation*}
where
\begin{equation*}
    R(u):=\frac{4\cos u}{1-\cos u}. 
\end{equation*}
Modulo Lemmas 1 and 2, it remains to show that $H>-4$. If $\cos u>0$, this follows because $r>0$.
If, finally, $\cos u<0$, then $R(u)>-2$ and hence, by Lemma 2, $H(u)=R(u) r(u)>-2\times1.25=-2.5>-4$. $\quad\Box$.

It remains to prove Lemmas 1 and 2.
Proof of Lemma 1: Recall that $h(u):=\sin2u-\sin u-u$. Note that $\sin2u-\sin u$ is $2\pi$-periodic in $u$, whereas $-u$ is decreasing in $u$. So, it suffices to show that $h<0$ on the interval $(0,2\pi)$. But this follows because
\begin{equation*}
    h'(u)=(\cos u-1)(4\cos u+3)\overset{\text{sign}}=-(4\cos u+3), 
\end{equation*}
so that the critical points of $h$ are easily found and considered. $\quad\Box$
Proof of Lemma 2: The inequality $r(u)<1.25$ can be rewritten as $b(u):=u+4\sin u>0$. Note that $\sin u$ is $2\pi$-periodic in $u$, whereas $u$ is increasing in $u$. So, it suffices to show that $b>0$ on the interval $(0,2\pi)$. But this follows because $b'(u)=1+4\cos u$, so that the critical points of $b$ are easily found and considered. $\quad\Box$

Remark 1: It follows from this proof that the lower bound $-4$ on $g(x)=G(a,u)$ in \eqref{1} can be replaced by the better lower bound $-2.5$. In fact, the best lower bound on $g(x)=G(a,u)$ in \eqref{1} is $\min_{u>0}H(u)=H(u_*)=-2.1289\dots$, where $u_*=3.5264\dots$. The minimization of $H(u)$ in $u>0$ can be effectively done by, say, the interval method, using the fact that $r$ is increasing on all intervals where $\cos<0$.
