In his 1926 paper Fermi states without further explanation that it follows from the Thomas-Fermi equation $$\frac{d^2\psi(x)}{dx^2}=\frac{\psi(x)^{3/2}}{\sqrt{x}},\label{1} \tag{1}$$ and boundary conditions $$\psi(0)=1,\quad\psi(\infty)=0\label{2}\tag{2},$$ that $$\int\limits_0^\infty{\frac{\psi^{5/2}(x)}{\sqrt{x}}}dx=-\frac{5}{7}\psi^\prime(0),\label{3} \tag{3}$$ where $\psi^\prime(x)=\frac{d\psi(x)}{dx}$. It seems Fermi considered this inference trivial. However, the proof given below and inspired by Kleinert, p. 429, although simple, is not at all trivial.

The Thomas-Fermi equation \eqref{1} can be considered as a Euler-Lagrange equation coresponding to the action principle $$ \delta S=0,\;\; S=\int\limits_0^\infty\left [\frac{1}{2}\left (\frac{d\psi}{dx}\right)^2+\frac{2}{5}\frac{\psi^{5/2}(x)}{\sqrt{x}}\right ]dx.\label{4}\tag{4} $$ The following infinitesimal deformation of the "coordinate" function $$\psi(x)\to \bar\psi(x)=\psi(\lambda x),\;\lambda=1+\epsilon,\;\epsilon\ll 1,$$ respects the boundary conditions \eqref{2} and changes the action functional \eqref{4} to $$\bar S=\int\limits_0^\infty\left [\frac{1}{2}\left (\frac{d\bar\psi(x)}{dx}\right)^2+\frac{2}{5}\frac{\bar\psi^{5/2}(x)}{\sqrt{x}}\right ]dx=\int\limits_0^\infty\left [\frac{\lambda}{2}\left (\frac{d\psi(y)}{dy}\right)^2+\frac{2}{5}\frac{\psi^{5/2}(y)}{\sqrt{\lambda y}}\right ]dy,$$ where $y=\lambda x$. Therefore, using $\lambda =1+\epsilon,\;\lambda^{-1/2}\approx 1-\epsilon/2$, we get $$\bar S =S+\epsilon \int\limits_0^\infty\left [\frac{1}{2}\left (\frac{d\psi(y)}{dy}\right)^2-\frac{1}{5}\frac{\psi^{5/2}(y)}{\sqrt{y}}\right ]dy,$$ and since we must have $\delta S=\bar S-S=0$ for any $\epsilon\ll 1$, we conclude $$\int\limits_0^\infty \left (\frac{d\psi(x)}{dx}\right)^2 dx=\frac{2}{5}\int\limits_0^\infty \frac{\psi^{5/2}(x)}{\sqrt{x}} dx.\label{5} \tag{5}$$ On the other hand, using \eqref{1}, \eqref{2} and integration by parts, we have $$ \begin{split} \int\limits_0^\infty \frac{\psi^{5/2}(x)}{\sqrt{x}} dx &=\int\limits_0^\infty \psi(x)\frac{d^2\psi(x)}{dx^2}dx\\ &=\left .\psi(x)\frac{d\psi(x)}{dx}\right |_0^\infty-\int\limits_0^\infty \left (\frac{d\psi(x)}{dx}\right)^2 dx\\ & =-\left .\frac{d\psi(x)}{dx}\right |_{x=0}-\int\limits_0^\infty \left (\frac{d\psi(x)}{dx}\right)^2 dx, \end{split}$$ and the Fermi result \eqref{3} immediately follows from \eqref{5}.

Is there a simpler proof of \eqref{3} that Fermi might have had in mind?