Something similar to the bizarre Koide formula? In 1981, Koide found the empirical relation,
$$\frac{m_e+m_\mu+m_\tau}{\big(\sqrt{m_e}+\sqrt{m_\mu}+\sqrt{m_\tau}\big)^2} = 0.666659\dots\approx \frac{2}{3}\tag1$$
where $m$ are the masses of the three leptons: electron, muon, and tau, approximately, $0.5, 105, 1776$. Later, about 2011, an identical relation was found,
$$\frac{m_c+m_b+m_t}{\big(\sqrt{m_c}+\sqrt{m_b}+\sqrt{m_t}\big)^2} = 0.666649\dots\approx \frac{2}{3}\tag2$$
where $m$ are the masses of the three heaviest quarks: charm, bottom, and top, approx $1290, 4370, 174100$. Some physicists dismiss these two as just numerology.
I was wondering how easy it was to find such relations so, using a small set of constants, found,
$$\frac{\pi+e+K}{\big(\sqrt{\pi}+\sqrt{e}+\sqrt{K}\big)^2} \approx 0.33376\dots\frac{1}{3}\tag3$$
where $K$ is the Khinchin constant.

Q: Let's make things difficult. Using three distinct constants $c_i$ that are a) well-known, b) transcendental (proven or not), and, c) like the Koide formula, the $c_i$ are not powers or multiples of each other, nor close to each other,$$\frac{c_1+c_2+c_3}{\sqrt{c_1}+\sqrt{c_2}+\sqrt{c_3}}\approx R$$ can you find a combination of $c_i$ that is nearer to some rational $R$ than $(1)$ or $(2)$? 

 A: Given your changes (i.e. wanting limit of $\frac23$) I'm giving a second answer. Again the problem boils down to what constants you use and the following results are from a exhaustive search of a limited number of constants and some fractions and powers of them.
Using the same function as in my first answer
$$f(x,y,z)=\frac{x+y+z}{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2}$$
the following combinations of constants give results close to $\frac23$.
Note: $A$ is the Glaisher–Kinkelin constant, $A=1.2824271291...$
$C_{10}$ is the Champernowne Number in base 10, $C_{10}=0.1234567891011121314151617...$
$G$ is the Catalan constant, $G=0.915965594177219...$
$$f\left(8,\frac{\pi^6}{64},(C_{10})^3\right)=0.66666688...$$
$$f\left(\frac{2\sqrt2}{27},7\sqrt7,\frac{\phi}{3}\right)=0.66666689...$$
$$f\left(\frac54,\left(\frac{A}{5}\right)^2,\frac{(C_{10})^2}{16}\right)=0.66666767...$$
$$f\left(7,\frac{G}{4},\frac{C_{10}}{4}\right)=0.66666801...$$
A: To take a random example, if $G$ is Catalan's constant and $\gamma$ the Euler-Mascheroni constant, then
$$ \dfrac{3 G^2+2 G+\pi^2-4 \pi}{\ln(\pi^2 \gamma G)} \approx 0.9999999413$$
A: Let $$f(x,y,z)=\frac{x+y+z}{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2}$$
Using $e$, Khinchin's constant and the golden angle in radians ($(3-\sqrt{5}\pi=2.39996...$) then the ratio is 0.33359021...
The reason for this is that $f(a,a,a)=\frac13$ so you just need to pick values which are close together.
