If the convolution with $\mid x\mid^{-1}$ is bigger, is the $L^1$ norm also bigger?

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I'm trying to understand the proof of the following statement:

Let $f_1,f_2\in L^1(\mathbb{R}^3)$ with $f_1(x), f_2(x)\geq 0$ for almost all $x$ in $\mathbb{R}^3$. For $i=1,2$, let $g_i=\mid x\mid^{-1} \ast f_i$. If $g_1(x)\geq g_2(x)$ for almost all $x$ in $\mathbb{R}^3$, then $\int_{\mathbb{R}^3} f_1\geq\int_{\mathbb{R}^3} f_2$.

In an article of E. Lieb (http://ergodic.ugr.es/statphys/bibliografia/lieb2.pdf, lemma 3.3) there is a proof which uses spherical averages and Newton Theorem (see Lieb, Loss, thm 9.7), but this would require $f_1-f_2$ to be radial symmetric, if I understand correctly, therefore I'm a little confused.

Thank you for any help.