A bound for a real rational function

Question

Let $r\geq 1, $ and $\theta$ be any real number. Then \begin{equation}\frac{1-r\cos\theta}{1+r^2-2r\cos\theta}\leq \frac{1}{2}.\end{equation} For, wehave $2\leq 1+r^2\implies2-2r\cos\theta\leq1+r^2-2r\cos\theta,$ which establishes the above inequality. The above inequality is sharp when $r=1,$ that is, when $r$ gets the least possible value. So for $r$ sufficiently larger than $1$, the gap between LHS and RHS widens. Is there any way of bettering this inequality by redefining the R.H.S by a function $f(r)$ where $f(1)=1/2?$ Kindly suggest me.

Answer 1

Hint: Multiplying both sides by $$1+r^2-2r\cos(\theta)$$ we get $$2-2r\cos(\theta)\le 1 +r^2-2r\cos(\theta)$$ Can you finish? $$1-2r\cos(\theta)+r^2=(r-1)^2+2r(1-\cos(\theta))\geq 0$$