Show that $(\frac{y-1}{y+1})^k\le \epsilon$, $0 \lt \epsilon \lt 1$, $y \ge 1,$ if $k \ge -\frac{1}{2}log(\epsilon)y$.
My idea is to derive the second inequality from the first.
taking $\log()$ on both sides, yields $k \log(\frac{y-1}{y+1}) \le \log (\epsilon)$ $\implies k \ge \frac{\log (\epsilon)}{log(\frac{y-1}{y+1})}$ because this estimate is the sharpest you can get. Unfortunaly, I dont the a way to conclude above.
Hints are appreciated. Greetings.
No, you shouldn't attempt to derive the second inequality from the first, given that you are required to do precisely the opposite! Continuing from where you stopped, you want to show that
$$k \ge -\frac y 2 \log \epsilon \implies k \ge \frac {\log \epsilon} {\log \dfrac {y-1} {y+1}}$$
and it is sufficient (but not necessary) to show that
$$-\frac y 2 \log \epsilon \ge \frac {\log \epsilon} {\log \dfrac {y-1} {y+1}}$$
or, equivalently,
$$\log \frac {y+1} {y-1} \ge \frac 2 y .$$
Consider the map $f : (1, \infty) \to \Bbb R$ given by $f(y) = \log \dfrac {y+1} {y-1} - \dfrac 2 y$. We want to show that $f \ge 0$.
We have
$$f' (y) = \frac {y-1} {y+1} \frac {(y-1) - (y+1)} {(y-1)^2} + \frac 2 {y^2} = \frac {-2} {(y-1)(y+1)} + \frac 2 {y^2} = \frac {-2} {y^2(y-1)(y+1)} \le 0$$
which shows that $f$ is decreasing on $(1, \infty)$, so that $f(y) \ge \lim \limits _{y \to \infty} f(y) = 0$, which is exactly what we wanted.
The above establishes the desired inequality for $y>1$. It is easy to check it by hand for $y=1$, whence it will hold true for $y \ge 1$ as desired.