How does $2\le1+2\cos x\le3$ become $\frac{2}{3}\le\frac{2}{1+2\cos x}\le1$?

Question

How does $2\le1+2\cos x\le3$ become $\frac{2}{3}\le\frac{2}{1+2\cos x}\le1$?

It seems that everything is being divided by $3$, as $2$ in the original inequality becomes $\frac{2}{3}$, and likewise, $3$ becomes $1$. But after dividing, we have:

$$\frac{2}{3} ≤ \frac{1 + 2 \cos x}{3} ≤ 1$$

where the middle term should be $\frac{2}{1 + 2 \cos x}$ instead of $\frac{1 + 2 \cos x}{3}$.

I know that the middle fraction gets flipped and multiplied by $2$. How do I use this to obtain the correct values of $\frac{2}{3}$ and $1$ in the new inequality?

Answer 1

Taking the reciprocal of our original inequality will swap the LHS and RHS around. Hence if, $$ 2 \leq 1 +2 \cos x \leq 3 $$ then $$ \frac{1}{3} \leq \frac{1}{1+2\cos x} \leq \frac{1}{2} $$ (provided that $\cos x \neq -1/2$). Finally, multiplying this expression by $2$ will give us the required inequality, $$ \frac{2}{3} \leq \frac{2}{1+2\cos x} \leq 1. $$

Answer 2

Because $1+2\cos{x}>0$ and $$\frac{2}{1+2\cos{x}}\leq1.$$ The second is similar: $$\frac{3}{1+2\cos{x}}\geq1$$ or $$\frac{2}{1+2\cos{x}}\geq\frac{2}{3}.$$

Answer 3

Think of it as two smaller inequalities:

$$2 ≤ 1 + 2 \cos x \Rightarrow \frac{2}{2} ≥ \frac{2}{1 + 2 \cos x} \tag{$\cos x \ne -\frac{1}{2}$}$$

as $a ≤ b \Rightarrow \frac{1}{a} ≥ \frac{1}{b}$.

The other inequality is similar: $$1 + 2 \cos x ≤ 3 \Rightarrow \frac{2}{1 + 2 \cos x} ≥ \frac{2}{3} \tag{$\cos x \ne -\frac{1}{2}$}$$

and combining the two inequalities gets us the desired answer.