I want to show that following: $$\left(\frac{n^2-1}{n^2}\right)^n\sqrt{\frac{n+1}{n-1}}\leq 1; ~~n\geq 2$$ and $n$ is an integer.
After some simplifications, I got left hand-side as $$LHS:\left(1-\frac{1}{n}\right)^{n-\frac{1}{2}} \left(1+\frac{1}{n}\right)^{n+\frac{1}{2}}$$ It is clear that the 1st term is less than 1, but I do not have any clue how I can show that multiplication is less than 1.
Can someone give me some hints?
Let
$$f(x) = \Big(\frac{x^2-1}{x^2}\Big)^x\sqrt{\frac{x+1}{x-1}}$$
Show that $f$ is monotonically increasing for $x\geq 2$ and that $f\rightarrow 1$ as $x\rightarrow\infty$.
Edit: Might be easier to show that $\log f$ is monotonically increasing and approaches zero.
Consider on $(-1,1)$ the function (with motivation $x=\frac1n$) $$ f(x)=\ln\bigl(1-x^2\bigr)+\frac x2\bigl(\ln(1+x)-\ln(1-x)\bigr)-\frac16\ln\bigl(1-x^4\bigr) $$ Then $$ f'(x)=-\frac{x}{1-x^2}+\frac12\bigl(\ln(1+x)-\ln(1-x)\bigr)+\frac16\frac{4x^3}{1-x^4} $$ where we see that $f(0)=0=f'(0)$. The next derivative is \begin{align} f''(x)&=-\frac{2x^2}{(1-x^2)^2}+\frac{2x^2}{1-x^4}+\frac83\frac{x^6}{(1-x^4)^2}\\ &=-\frac{4x^4}{(1-x^4)(1-x^2)}+\frac83\frac{x^6}{(1-x^4)^2}\\ &=-\frac43·\frac{3x^4+x^6}{(1-x^4)^2} \end{align} which is always negative for $x\ne 0$. Then for $x>0$ the linear Taylor polynomial with quadratic remainder term gives $$ f(x)=\frac12f''(\theta x)x^2<0 $$ and taking the exponential of $f(x)<0$ gives $$ \bigl(1-x^2\bigr)·\left(\frac{1+x}{1-x}\right)^{\frac x2}<\bigl(1-x^4\bigr)^{\frac16}. $$ Replacing $x=\frac1n$ and taking the $n$th power of the inequality results in the requested inequality, $$ \left(1-\frac1{n^2}\right)^n·\sqrt{\frac{n+1}{n-1}}<\left(1-\frac1{n^4}\right)^{\frac n6}<1. $$
