show this $(f(x))''+2\ge 0$,for any real numbers, where $f(x)=\frac{x^2-1}{x^{2n}-1}$ When I did a question today, I turned to the following question
let $n$ be postive integer,and  $x\in R$, let $f(x)=\dfrac{x^2-1}{x^{2n}-1}$,show this
$$(f(x))''+2\ge 0 \tag{1}$$
For example $n=2$
then
$$(f(x))''+2=\dfrac{2x^2(x^4+3x^2+6)}{(x^2+1)^3}\ge 0$$
$n=3$,then
$$(f(x))''+2=\dfrac{2x^4(x^8+3x^6+6x^4+17x^2+15)}{(x^4+x^2+1)^3}\ge 0$$
and when $n=4$,see links1
$n=5$ see  links2,
$n=6$ see links3
always hold,so How to prove for $(1)$,maybe this usefull:
\begin{align*}(f(x))''+2&=(f(x)+x^2)''=\left(\dfrac{x^{2n+2}-1}{x^{2n}-1}\right)''\\
&=\dfrac{x^{2n-2}(4n^2(x^2-1)(x^{2n}+1)-2n(3x^2+1)(x^{2n}-1)+2x^2(x^{2n}-1)^2)}{(x^{2n}-1)^3}\end{align*}see links4

As stated above in a comment, we can write
$$f(x)=\frac{1}{1+x^2...+x^{2n-2}}=\frac{1}{g(x)}$$
which yields that
$$f''(x)+2=\frac{2g'^2-gg''+2g^3}{g^3}$$
so we only have to decide the sign of the numerator, given that $g(x)>0$ for all x. Since $f(x)=f(-x)$ it also follows that the second derivative is even, which allows us to only examine $x\geq 0$. It seems that the coefficients of the numerator are explicitly positive, so let us see how we can prove this. First, write out the expressions for $g'^2 , gg''$:
$$2g'^2=\left(\sum_{r=0}^{n-1}\sum_{l=0}^r+\sum_{r=n}^{2n-2}\sum_{\ell=r-n+1}^{r}\right)8\ell(r-\ell)x^{2r-2}$$
$$g''g=\left(\sum_{r=0}^{n-1}\sum_{l=0}^r+\sum_{r=n}^{2n-2}\sum_{\ell=r-n+1}^{r}\right)2\ell(2\ell-1)x^{2r-2}$$
Subtracting these two we see that
$$2g'^2-gg''=-\sum_{r=0}^{n-1}x ^{2r-2}r(1+r)+\sum_{r=n}^{2n-2}[4(r-n)^2+6r-5n+1]x^{2r-2}$$
Note that in the 2nd sum all the coefficients are positive and in the 1st sum negative. We will prove that the expansion of $2g^3$ exactly cancels out the $n-1$ smallest powers (note that $r=0$ does not contribute but is included for convenience).
After some manipulations we see that
$$g^3=\sum_{r=0}^{3n-3}x^{2r}\sum_{s=\max\{0,r-n+1\}}^{r}\sum_{m=\max\{0,s-n+1\}}^{s}1$$
Here, all coefficients are manifestly positive as well. Now for $0\leq r\leq n-2$ we immediately see that after performing the sum and a change of index:
$$2g^3=\sum_{r=0}^{n-2}x^{2r}\sum_{s=0}^r\sum_{m=0}^s 1+...=\sum_{r=0}^{n-1}x^{2r-2}r(1+r)+\sum_{r=n-1}^{3n-3}A_{nr}x^{2r}~, ~A_{nr}>0$$
which finally shows that the first $n-1$ coefficients are cancelled and the remaining expression is a polynomial in $x^2$ of degree $3(n-1)$ with positive coefficients and therefore is positive for all $x>0$. At $x=0$ evidently, $f''(0)+2=0$ and hence $f''(x)+2\geq 0$ for all $x$, with a zero of order $2n-2$ at the origin.
