Bounding higher derivatives of $f(x) = 1/(1+x^2)^r$

Question

Let $r\in \lbrack 0,\infty)$. Define $f(x) = 1/(1+x^2)^r$. It would seem to be the case that $$|f^{(k)}(x)|\leq \frac{2r \cdot (2r+1) \dotsb (2r + k-1)}{(1+x^2)^{r + k/2}}$$ for all even $k\geq 0$. What would be a clean, short proof?

Note that $$\left(\frac{1}{(1+x^2)^r}\right)^{(k)} = \frac{P_k(x)}{(1+x^2)^{r+k}},$$ where $P_k\in \mathbb{Z}\lbrack x\rbrack$ is given by the recurrence formula $$P_k(x)=1,\;\;\;\;\;\;P_{k+1}(x) = P_k'(x) (1+ x^2) + 2 (r+k)\cdot x P_k(x).$$ It should not be too hard to work out all the coefficents of $P_k(x)$; the leading terms are $$(-1)^{k+1} \cdot 2r \cdot (2r+1) \dotsb (2r + k-1)\cdot \left(x^k - \binom{k}{2} \frac{x^{k-2}}{2 r + 1} + \dotsc\right).$$ The desired bound should then follow after some work -- but I am hoping for a better proof.

Answer 1

Well, $$ (a^2+x^2)^{-r}=(x+ai)^{-r}\cdot (x-ai)^{-r}. $$ Differentiate this $k$ times, we get $$ \left((a^2+x^2)^{-r}\right)^{(k)}=k!\sum_{s=0}^k {-r\choose s}{-r\choose k-s} (x+ai)^{-r-s}(x-ai)^{-r-(k-s)}. $$ Estimate this by the triangle inequality, you get some constant times $(x^2+a^2)^{-r-k/2}$. But for $a=0$ we have equality (because the signs of all summands are equal to $(-1)^k$), thus the constant is $$ (-1)^kx^{2r+k}\left(x^{-2r}\right)^{(k)}=2r(2r+1)\ldots(2r+k-1). $$