Consider the following function $$f(t) := \begin{cases} |t|,~~ &\text{if $|t| > 1$} \\ 1,~~ &\text{if $|t| \leq 1$}. \end{cases}$$
I'm trying to find a reasonably accurate constant $M > 0$ such that $$ \int_{-\infty}^\infty \frac{f(s)^2}{f(t)^2f(s-t)^2}\; dt ~\leq~ M$$ for all $s \in \mathbb{R}$.
If $|s| \leq 1$ then $$\int_{-\infty}^\infty \frac{f(s)^2}{f(t)^2f(s-t)^2}\; dt ~\leq~ \int_{-\infty}^{-1}\frac{1}{t^2}\; dt + \int_{-1}^1 \; dt + \int_{1}^\infty \frac{1}{t^2}\; dt ~\leq~ 5$$
But I don't know how to tackle the case $|s|> 1$. Any hint ?
True though maybe not optimally helpful: Mathematica says that the exact value of the integral is $$ \begin{cases} \frac83 & s=0 \\ \frac{-s^4+4 s^3-2 s^2+4 s-4 \log (s+1)}{s^3} & 0<s\leq 1 \\ \frac{-s^4+4 s^3-2 s^2+4 s-4 \log (s+1)}{s} & 1<s\le2 \\ \frac{4 (2 s+\log (s-1)-\log (s+1))}{s} & s>2,\end{cases} $$ and so the best possible constant is $8$.