Lower bound via Sobolev Inequality Let $ f \in L^1(\mathbb{R}^3) $ be such that $ \int f =1 $ and $ \int \tfrac{f}{\|x\|} < \infty $.
I want to show that
\begin{equation}\label{key}
 27 \left( \dfrac{\pi}{2} \right)^4 \int f^3 \ge \left( \int (\dfrac{1}{\| x\|} - \dfrac{1}{3} ) f\right)^3. 
\end{equation}
Now, it is clear to me that the integrals converge by the assumptions given. The hint given is to use H/"older inequality.
Now, the most obvious step would be to choose $ p=3, q=3/2 $, and do Holder on $ f $. However, I run into the issue that $ (\tfrac{1}{\|x\|}-\tfrac{1}{3}) $ is not integrable.
I was wondering if this was the right approach? Or do I need to take a positive part first?
 A: The claimed inequality is false
I assume the functions are real-valued. Taking the cube root (which is order preserving on the reals), your claimed inequality is equivalent to
$$ 3 \cdot \left(\frac\pi2\right)^{4/3} \left(\int f^3 \right)^{1/3} + \frac13 \int f \geq \int \frac{1}{|x|} f $$
We note that the assumption that $\int f = 1$ is inconsequential, as long as $\int f > 0$. For given any $f$ with positive total integral, you can set $g = (\int f)^{-1} f$ to get that $\int g = 1$. Applying the inequality to $g$ we can restore to an inequality for $f$ by multiplying throughout by $\int f$.
I claim in fact that the result cannot hold for any sets of constants. More precisely

Claim For any $A, B > 0$, there exists $f\in L^1$ with $\frac{1}{|x|}f$ integrable, and $\int f > 0$, such that
$$ A (\int f^3)^{1/3} + B \int f < \int \frac{1}{|x|} f $$

To prove this claim, we will consider the three parameter family of functions $f_{\mu,\kappa,\rho}$, with $\mu,\rho > 0$ and $\kappa > 1$
$$ f_{\mu,\kappa,\rho}(x) = \begin{cases} -\mu & |x| \leq \rho \\
1 & |x| \in [\kappa\rho, 2\kappa\rho] \\
0 & \text{otherwise} \end{cases} $$
A direct computation yields that
$$ \int f_{\mu,\kappa,\rho} = \frac43 \pi \rho^3 \left( 7 \kappa^3 - \mu\right) $$
$$ \int \frac{1}{|x|} f_{\mu,\kappa,\rho} = 2\pi \rho^2 \left( 3 \kappa^2 - \mu\right) $$
$$ \int f^3_{\mu,\kappa,\rho} = \frac43 \pi \rho^3 \left( 7 \kappa^3 - \mu^3 \right) $$
If you set $\mu = \kappa^{3/2}$, then there exists $\kappa_0$ such that for all $\kappa > \kappa_0$,
$$ 3\kappa^2 - \mu > \kappa^2, \qquad 7 \kappa^3 - \mu^3 < 0. $$
Now let $A, B > 0$ be given. Fix $\kappa > \kappa_0$ and $\mu = \kappa^{3/2}$.
Our goal is to find $\rho$ such that
$$ A (\int f_{\mu,\kappa,\rho}^3)^{1/3} + B \int f_{\mu,\kappa,\rho} < \int \frac{1}{|x|} f_{\mu,\kappa,\rho} $$
Our choice of $\kappa$ and $\mu$ guarantees that the first term is negative, and hence it suffices to find $\rho$ such that that
$$ B\int f_{\mu,\kappa,\rho} < \int \frac{1}{|x|} f_{\mu,\kappa,\rho} $$
or
$$ B \frac43 \pi \rho^3 \left( 7 \kappa^3 - \mu\right) < 2\pi \rho^2 \left( 3 \kappa^2 - \mu\right) $$
or, simplifying, it suffices to show that
$$ 2 B \rho \left( 7 \kappa^3 - \mu\right) < 3 \left( 3 \kappa^2 - \mu\right) $$
Our choice of $\kappa$ and $\mu$ guarantees that the RHS is at least $3 \kappa^2$, and the LHS is at most $14 B \rho  \kappa^3$. And so if we choose
$$ \rho < \frac{3}{14 B \kappa} $$
we obtain the desired counterexample.
Remark
This counterexample strongly depends on the fact that I am taking at face value that the cubic integral is of $f^3$ and not $|f|^3$. This allows me to take the cubic integral out of the action and avoid any Holder like controls.
Indeed, if we make the additional assumption that $f$ is a non-negative function, then we would have that
$$ \int \left(\frac1{|x|} - \frac13\right) f \leq \int_{|x| < 3} \left(\frac1{|x|} - \frac13\right) f \leq \tilde{C} \|f\|_{L^3} $$
through Holder inequality, where
$$ \tilde{C} = \left( 4\pi \int_0^3 (1/r - 1/3)^{3/2} r^2 ~dr \right)^{2/3} = \left( \frac{\pi^2 3^{3/2}}{4} \right)^{2/3} $$
the evaluation performed by trig substitution. And you see that this is the exact constant claimed in the question. So I strongly suspect that the question is intended only for $f \geq 0$, and not general $f$ as stated.
