How to solve the following simultaneous equations?

Question

It doesn't look like it can be solved by the normal substitution method:

It is given in a paper and the solution is given as $m \geq 2 + \sqrt(3)$.

Answer 1

As I said in a comment, it looks like two inequalities for two possible cases depending on some threshold value. Let's look at them one at a time. Before we start, some observations from the original paper: $\theta_{12}$ is a positive angle; the parameter $m$ satisfies $m\ge2$; and they cautiously say that "we could have $m_{\text{min}}\ge2+\sqrt{3}$" (note "could" rather than "is").

Case 1. Note that $\theta_{12}$ is a common factor in all terms of the inequality. Since it's a positive number, we can divide both sides of the inequality by $\theta_{12}$: $$\frac{1}{m-1}+\frac{1}{m+1}\le\frac{m-1}{m+1}.$$ Since $m\ge2$, we know that $m-1>0$ and $m+1>0$, so we can multiply both sides by $(m-1)(m+1)$: $$(m+1)+(m-1)\le(m-1)^2.$$ Rearranging the terms and simplifying, we end up with a pretty standard quadratic inequality $$m^2-4m+1\ge0,$$ whose solutions is $$m\in(-\infty,2-\sqrt{3}]\cup[2+\sqrt{3},+\infty).$$ Since $m\ge2$ and they are looking for a minimal value of $m$, that's why from this answer it's reasonable (in the context) to say that "we could have $m_{\text{min}}\ge2+\sqrt{3}$".

Case 2. Note that $m+1$ is the same denominator in all terms of the inequality. Since it's a positive number, we can multiply both sides of the inequality by $m+1$ to obtain: $$(2\pi-\theta_{12})+\theta_{12}\le(m-1)\theta_{12},$$ which simplifies, using the fact that $m-1$ is positive, to: $$\theta_{12}\ge\frac{2\pi}{m-1}. \tag{1}\label{sol}$$ This case is a little trickier. Remember that we're considering the case when $$\theta_{12}>\frac{m-1}{m}\pi. \tag{2}\label{con}$$

To make sure that a value of $\theta_{12}$ that satisfies the constraint \eqref{con} will satisfy the desired inequality \eqref{sol}, we should require that $$\frac{m-1}{m}\pi\ge\frac{2\pi}{m-1}.$$ We can divide both sides by the common factor of $\pi$. Rearranging and simplifying further, we (interestingly enough!) end up with the same quadratic inequality $$m^2-4m+1\ge0,$$ thus obtaining the same solution for this case too.