Prove that $\sup_y\int_{-\infty}^\infty\frac{dx}{(1+x^2)|x-y|}$ is obtained at $y=0$ [closed]

How to show that for a constant $0<\delta<1$, if $f$ is even, $f'(x)<0$ for $x>0$ and decays rapidly at infinity, then the supremum $\sup_y\int_{-\infty}^\infty\frac{f(x)}{|x-y|^\delta}\,dx$ is obtained at $y=0$?

Idea of proof: if $y>x$, then $\frac{d}{dy}\frac{1}{|x-y|^\delta}=\frac{d}{dy}\frac{1}{(y-x)^\delta}=-\frac{d}{dx}\frac{1}{(y-x)^\delta}=-\frac{1}{dx}\frac{1}{|x-y|^\delta}$ and if $y<x$ then $\frac{d}{dy}\frac{1}{|x-y|^\delta}=\frac{d}{dy}\frac{1}{(x-y)^\delta}=-\frac{d}{dx}\frac{1}{(x-y)^\delta}=-\frac{1}{dx}\frac{1}{|x-y|^\delta}.$

So since this holds for both cases we have $\frac{d}{dy}\int_{-\infty}^\infty\frac{f(x)}{|x-y|^\delta}\,dx=\int_{-\infty}^\infty-\frac{d}{dx}\frac{1}{|x-y|^\delta}f(x)\,dx=\int_{-\infty}^\infty\frac{1}{|x-y|^\delta}f'(x)\,dx.$ Since $f$ is even about $0$, we have that $f'(x)$ is odd about $0$, and thus $y=0$ implies that $\frac{d}{dy}\int_{-\infty}^\infty\frac{f(x)}{|x-y|^\delta}\,dx=0$ and so $y=0$ is a point of inflection.

How do I prove that this a maximum not a minimum point?