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?
 A: Hi @max We can use Binomial series as we have :
$$\left(1-\frac{2}{x^{2}}\right)\leq \left(\frac{\left(x-1\right)}{x+1}\right)^{\frac{1}{x}}=\left(\frac{\left(x-1\right)}{x+1}\right)^{1+\frac{1}{x}-1}$$
So at order two we have :
$$\left(\frac{\left(x-1\right)}{x+1}\right)^{-1}\left(1+\left(\frac{\left(x-1\right)}{x+1}-1\right)\left(1+\frac{1}{x}\right)+\frac{1}{2}\left(\frac{\left(x-1\right)}{x+1}-1\right)^{2}\left(1+\frac{1}{x}\right)\left(\frac{1}{x}\right)\right)\leq \left(\frac{\left(x-1\right)}{x+1}\right)^{\frac{1}{x}}$$
But :
$$\left(\frac{\left(x-1\right)}{x+1}\right)^{-1}\left(1+\left(\frac{\left(x-1\right)}{x+1}-1\right)\left(1+\frac{1}{x}\right)+\frac{1}{2}\left(\frac{\left(x-1\right)}{x+1}-1\right)^{2}\left(1+\frac{1}{x}\right)\left(\frac{1}{x}\right)\right)=\left(1-\frac{2}{x^{2}}\right)$$
We are done !


Some idea fo someone who wants to show Angelo problem :
For $x\geq 2$ we have :
$$\cos\left(\frac{2}{x}\right)-\cos\left(\frac{2}{\sqrt{x}}\right)\geq 2\left|\frac{x-1}{x^{2}+\frac{2}{3}+\frac{1}{3}x}\right|\geq \left(\frac{x-1}{x+1}\right)^{\frac{1}{x}}-\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)^{\frac{1}{\sqrt{x}}}$$
Remains to show :
$$\lim_{n\to \infty}\cos\left(\frac{2}{x^{2n}}\right)-\left(\frac{x^{2n}-1}{x^{2n}+1}\right)^{\frac{1}{x^{2n}}}$$
A: Taking logarithms leads to
$$
x\log\left(1-\frac2{x^2}\right)\lt\log(x-1)-\log(x+1)
$$
and thus
$$
\log\left(1-\frac2{x^2}\right)\lt\frac1x\left(\log\left(1-\frac1x\right)-\log\left(1+\frac1x\right)\right)\;.
$$
So we want to show that
$$
\log\left(1-2z^2\right)\lt z(\log(1-z)-\log(1+z))
$$
for $0\lt z\lt\frac12$. Expanding both sides into power series leads to
$$
-\sum_{k=1}^\infty\frac{\left(2z^2\right)^k}k\lt-z\left(\sum_{j=1}^\infty\frac{z^j}j-\sum_{j=1}^\infty\frac{(-z)^j}j\right)
$$
and thus
$$
\sum_{k=1}^\infty\frac{2^k}kz^{2k}\gt\sum_{k=1}^\infty\frac2{2k-1}z^{2k}\;.
$$
The claim follows since the terms for $k=1$ are equal and all other terms are larger on the left. This in fact holds for $0\lt z\lt\frac1{\sqrt2}$ and thus for $x\gt\sqrt2$.
A: 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 $x>\sqrt 2$.)
