Proving that ${(1-\frac{2}{x^2})}^x < \frac{x-1}{x+1}$ for any $x > 2$. Proof that ${\left(1-\dfrac{2}{x^2}\right)}^x\!< \dfrac{x-1}{x+1}$ for any $x > 2$.
any ideas?

Let $f(x)=\ln\frac{x-1}{x+1}-x\ln(1-2/x^2)$. It suffices to show that $f(x)>0$ for $x>2$, as $\ln$ is an increasing function. Using the series for $\ln(1+a)=a-a^2/3+a^3/3-a^4/4+\dots$, we get the asymptotic expansion
$$
\ln \frac{x-1}{x+1}=\ln \left(1-\frac1x\right)-\ln \left(1+\frac1x\right)=-\frac{2}{x}-\frac{2}{3x^3}-\frac{2}{5x^5}-\frac{2}{7x^7}-\dots
$$
and
$$
-x\ln\left(1-\frac2{x^2}\right)=\frac2x+\frac{2^2}2 \frac{1}{x^3}+\frac{2^3}3 \frac{1}{x^5}+\frac{2^4}4 \frac{1}{x^7}+\dots
$$
hence
$$
f(x)=\sum_{k=1}^\infty \frac 1{x^{2k+1}}\left(\frac{2^{k + 1}}{k + 1} - \frac2{2k+1}\right)>0
$$
since
$$
\frac{2^{k + 1}}{k + 1} - \frac2{2k+1}=2\frac{2^k(2k+1)-(k+1)}{(k+1)(2k+1)}>0.
$$
(Note that the inequality $f(x)>0$ holds for all $x>\sqrt 2$.)
