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 $g(x)>-4$ (an example is attached); any hints to prove this analytically?


 A: The following consists of partial results and ideas. The substitutions $\beta x\mapsto x$ and $\alpha\beta\mapsto \alpha$ shows that one may assume without loss of generality that $\beta=1$. Then for $x>0$ and $\alpha\in {\mathbb R}$, one needs to show that $$f(x)=\frac{\alpha\sin x}{x}+4\cos x<0\qquad (1)$$ implies $$g(x)=\frac{\alpha\cos x}{x^2}+\frac{\alpha\cos 2x}{x^2}-\frac{\alpha\cot(x/2)}x-\frac{4\sin 2x}x+4>0.\qquad (2)$$ The problem can be proven by a case-by-case analysis of the following situations: The behavior of $f,g$ (a) when $x\rightarrow 0^+$, (b) when $x\rightarrow \infty,$ (c) when $x/2$ is near the points where $\cot$ diverges, and of course (d) when $|\alpha|$ is small or large.
Trivial case. $\alpha=0.$ One needs to show that $$4\cos x<0,x>0\Rightarrow -\frac{4\sin 2x}x+4>0\Leftrightarrow x>\sin 2x.$$ But the LHS implies $x>\frac {\pi}2>1\geq \sin 2x,$ hence the RHS.
Case when $x\geq \max(8,|2\alpha|^{\frac 12}).$ Then $$g(x)=\left(1+\frac{\alpha\cos x}{x^2}\right)+\left(1+\frac{\alpha\cos 2x}{x^2}\right)+\left(1-\frac{4\sin 2x}x\right)+\left(1-\frac{\alpha\cot(x/2)}x\right)$$ $$\geq \frac 1 2+\frac 1 2+\frac 1 2+1-\frac{\alpha\cot(x/2)}x.$$ So it suffices to prove that if (1) holds, then $$\frac 5 2-\frac{\alpha\cot(x/2)}x=\frac 5 2-\frac {\alpha}{xt}>0,\qquad (3)$$ where $t=\tan(x/2).$ To do this, use rational parametrization of $\sin x$ and $\cos x$ in terms of $t$, i.e. $\sin x=\frac{2t}{1+t^2},\cos x=\frac{1-t^2}{1+t^2}.$ Then (1) can be rewritten as $$2\alpha t+4x(1-t^2)<0$$ $$\Leftrightarrow t^2-\frac{\alpha}{2x}t-1>0,\qquad (4)$$ which is equivalent to $$t>\frac{\alpha}{4x}+\sqrt{\left(\frac{\alpha}{4x}\right)^2+1}~~{\rm or~}t<\frac{\alpha}{4x}-\sqrt{\left(\frac{\alpha}{4x}\right)^2+1},\qquad (5)$$ where $t=\tan(x/2).$ Now if $\alpha>0$ and $t<0$, (3) clearly holds. If $\alpha>0$ and $t>0$, then from the first part of (5) it follows that $$t>\frac {\alpha}{2x}\Rightarrow 1>\frac{\alpha}{2xt}~(t>0~{\rm in~this~case})$$
$$\Rightarrow 2-\frac{\alpha}{xt}>0$$ $$\Rightarrow \frac 5 2-\frac{\alpha}{xt}>0,$$ hence (3) holds. On the other hand, if $\alpha<0$ and $t>0$, then (3) clearly holds. If $\alpha<0$ and $t<0$, then from the second part of (5), $$t<\frac{\alpha}{2x}$$
$$\Rightarrow 1>\frac{\alpha}{2xt}$$
$$\Rightarrow 2-\frac{\alpha}{xt}>0$$ $$\Rightarrow \frac 5 2-\frac{\alpha}{xt}>0,$$ concluding the proof of (3).
Other cases. (Not an exhaustive list). Plan of attack is briefly described.
(a) For $|\alpha|$ large, start by showing the following lemma.
Lemma. For $x>0$, one has $$\sin x(\cos x+\cos 2x-x\cot(x/2))<0$$ if $\sin x\neq 0.$ (The condition was found by ignoring terms without $\alpha$ in (1) and (2).)
(b) When $x\rightarrow 0^+,$ the limiting behavior of (1) and (2) will give clues on how to prove it in this case.
(c) When $x,\alpha$ are both bounded, use numerical methods if necessary.
Hope this helps.
