Prove or disprove that $ \min\Bigg\{\frac{2x(1+\cos\theta)}{x+y},\frac{2x(1-\cos\theta)}{x-y}\Bigg\}\leq 2.$

Question

Let $x>0$, $\theta\in\mathbb{R}$, and $y\in\mathbb{R}$ such that $|y|<x$. Prove or disprove by counterexample that \begin{equation} \min\Bigg\{\frac{2x(1+\cos\theta)}{x+y},\frac{2x(1-\cos\theta)}{x-y}\Bigg\}\leq 2. \end{equation} I faced it in a research problem and confirmed it numerically.

Answer 1

Let $$\min\left\{\frac{2x(1+\cos\theta)}{x+y},\frac{2x(1-\cos\theta)}{x-y}\right\}=k.$$ Thus, $$\frac{2x(1+\cos\theta)}{x+y}\geq k$$ and $$\frac{2x(1-\cos\theta)}{x-y}\geq k$$ or $$1+\cos\theta\geq k\left(\frac{1}{2}+\frac{y}{2x}\right)$$ and $$1-\cos\theta\geq k\left(\frac{1}{2}-\frac{y}{2x}\right),$$ which after summing gives $$2\geq k$$ Can you end it now?