The solution of this problem can be obtained using an alternative approach. A well known property of triangles is that there exist points that minimize the sum of the distances to the vertices. These are usually called Fermat points or isogonal centers. If all angles of the triangle are $<120^o$, then the first Fermat point is inside the triangle and is the point from which each side subtends an angle of $120^o$. Geometrical methods allow to identify this point quite easily in any triangle. If the triangle is equilateral, the point trivially coincides with the center of the triangle.
The fact that the center of an equilateral triangle minimizes the sum of the distances to the vertices can also be proved directly by geometrical considerations. It is known that, in an equilateral triangle, the sum of the distances to the sides, drawn perpendicularly from any point interior to the triangle, is constant and equal to the height (this is easily shown if we observe that these distances are the heights of the three smaller triangles obtainable by connecting the point to the vertices and whose bases are the sides, since these triangles have constant total area and constant equal bases). Now let us consider the equilateral triangle $ABC$ with center $O$ described in the question, and let us trace through $A$ a line parallel to $BC$, through $B$ a line parallel to $AC$, and through $C$ a line parallel to $AB$. This creates a larger equilater triangle $DEF$ that contains the triangle $ABC$. The sum $OA+ OB+ OC=S_o$ can be seen as the sum of the distances from $O$ to the sides of the larger equilateral triangle $DEF$. Therefore, if we consider another point $K$ inside the triangle, the sum of the distances from $K$ to the sides of the triangle $DEF$ is still equal to $S_o$. If we connect $K$ with $A$, $B$, and $C$, we have that the sum $KA+KB+KC=S_K$ is clearly larger than the sum of the distances from $K$ to the sides of the triangle $DEF$, which in turn is equal to $S_o$. So, moving from the center $O$ to the point $K$, we have $S_K>S_o$, i.e. the sum of the distances to the vertices $A, B, C$ increases.
Now let us focus on the sum $S_K=KA+KB+KC$. To estimate it, it is convenient to use polar coordinates. Setting the origin in $O$, the coordinates of a point $K$ whose distance from $O$ is $2\rho$ are ($2\rho \cos\alpha, 2\rho \sin\alpha$). In this way, we can explore how $S_K$ varies by changing the angle $\alpha$, that is to say by considering all points identified by rotating a segment of length $2\rho$ around the origin $O$. Taking into account the coordinates of $A, B, C$ already calculated and reported in the question, we can then write the sum of the distances from the points $A$, $B$, $C$ as
$$S_K(\alpha)=\sqrt{4 \rho^2-4 \rho(\cos\alpha-\frac {\sqrt{3}\sin\alpha}{3})+\frac{4}{3}}+\sqrt{4\rho^2+4\rho(\cos\alpha+\frac {\sqrt{3}\sin\alpha}{3})+\frac{4}{3}}+\sqrt{4\rho^2-\frac {8\rho \sqrt{3}\sin\alpha}{3}+\frac{4}{3}}$$
Now if we keep $\rho$ fixed and calculate the derivative of the last function for $\alpha$, we get
$$S_K'(\alpha)=\frac {2\rho (\frac{ \sqrt{3}\cos\alpha}{3}+\sin\alpha)} {\sqrt{4 \rho^2-4 \rho(\cos\alpha-\frac {\sqrt{3}\sin\alpha}{3})+\frac{4}{3}}} +\frac {2\rho (\frac{ \sqrt{3}\cos\alpha}{3}-\sin\alpha)} {\sqrt{4 \rho^2+4 \rho(\cos\alpha+\frac {\sqrt{3}\sin\alpha}{3})+\frac{4}{3}}} -\frac{ 4 \sqrt{3}\cos\alpha} { 3 \sqrt{4\rho^2-\frac {8\rho\sqrt{3}\sin\alpha}{3}+\frac{4}{3}}} $$
Thanks to the simmetry of the problem, we can focus only on the range $(-\frac{\pi}{2}, \frac{\pi}{2})$. In this interval, the derivative has four zeros in $\alpha=\pm \frac{\pi}{2}$ and $\alpha=\pm\frac{\pi}{6}$. Accordingly, for these values the terms of the derivative reduce to
$$S_K'(\frac{\pi}{2})=S_K'(-\frac{\pi}{6})=\frac {2\rho} {\sqrt{4 \rho^2+\frac {4 \rho\sqrt{3}\sin\alpha}{3}+\frac{4}{3}}} -\frac {2\rho} {\sqrt{4 \rho^2+\frac {4 \rho\sqrt{3}\sin\alpha}{3}+\frac{4}{3}}} =0$$
$$S_K'(-\frac{\pi}{2})=S_K'(\frac{\pi}{6})=\frac {2\rho} {\sqrt{4 \rho^2-\frac {4 \rho\sqrt{3}\sin\alpha}{3}+\frac{4}{3}}} -\frac {2\rho} {\sqrt{4 \rho^2-\frac {4 \rho\sqrt{3}\sin\alpha}{3}+\frac{4}{3}}} =0$$
The derivative is negative in the intervals $-\frac{\pi}{2}<\alpha<\frac{\pi}{6}$ and $\frac{\pi}{6}<\alpha<\frac{\pi}{2}$, and positive in the interval $-\frac{\pi}{6}<\alpha<\frac{\pi}{6}$. Thus, $\alpha=-\frac{\pi}{2}$ and $\alpha=\frac{\pi}{6}$ are maximum points of $S_K$, whereas $\alpha=-\frac{\pi}{6}$ and $\alpha=\frac{\pi}{2}$ are minimum points of $S_K$. It is also straightforward to show that the values of $S_K$ in the two maximum points are equal, and so are the values in the two minimal points. To calculate these maximal and minimal values of $S_K$, by simplicity we can then restrict our analysis only to the cases $\alpha=-\frac{\pi}{2}$ and $\alpha=\frac{\pi}{2}$.
These results are illustrated in the figure below:
The figure illustrates that, considering a point $K$ along the circumference obtained by rotating a segment of length $2\rho$ around the origin $O$, the minimal value of $S_K$ is obtained by placing the point $K$ on the segment $OC$ (corresponding to $\alpha=\frac{\pi}{2}$), whereas the maximal value of $S_K$ is obtained by placing the point in the opposite direction, nearest to the side $AB$ (corresponding to $\alpha=-\frac{\pi}{2}$) (the other points of maximum and minimum are not shown in the figure). For a given $\rho$, these maximal and minimal values ($S_{Kmax}$ and $S_{Kmin}$, respectively) can be obtained by substituting $\alpha=-\frac{\pi}{2}$ and $\alpha=\frac{\pi}{2}$ in the expression giving $S_P$. This leads to
$$S_{Kmax}=\frac{2 \sqrt{3}}{3}+2\rho+2\sqrt{(\frac{\sqrt{3}}{3}-2\rho)^2+1}$$
$$S_{Kmin}=\frac{2 \sqrt{3}}{3}-2\rho+2\sqrt{(\frac{\sqrt{3}}{3}+2\rho)^2+1}$$
Note that these values coincides for $\rho=0$, where they correspond to $2\sqrt{3}$ (this is the case in which the point $K$ coincides with the origin $O$ and then $S_K$ reduces to three times the distance between $O$ and any vertex $A, B, C$). Also note that, for $2\rho=\frac{2 \sqrt{3}}{3}$, we get $S_{Kmin}=4$ (this is the case in which the point $K$ coincides with the vertex $C$ and $S_K$ reduces to the sum of the two sides $CB$ and $CA$). Lastly, for $2\rho=\frac{ \sqrt{3}}{3}$, we get $S_{Kmax}=2+\sqrt{3}$ (this is the case in which the point $K$ is the central point of the side $AB$ and $S_K$ reduces to the sum of the side $AB$ and the height $KC$).
Now we can prove the geometric inequality described in the OP. Consider a point $P$ inside the triangle $ABC$ whose distance from the origin $O$ is $\rho$, and a second point $Q$ inside the triangle $ABC$ whose distance from the origin $O$ is $2\rho$. In analogy with $S_K$, let us call $S_P$ and $S_Q$ the sums of the distances from $P$ and $Q$ to the vertices $A,B,C$, respectively. The calculations used above to compute $S_K$ and its maximum and minimum points can be directly applied to $S_Q$. The same calculations are also valid for $S_P$, with the only difference that the distance to the origin $O$ (and then the radius of the circumference along which $P$ can rotate) is halved as compared to $Q$. The minimal possible difference between $S_Q$ and $S_P$ occurs when the positions of the points $Q$ and $P$ correspond to $S_{Qmin}$ and $S_{Pmax}$. Because we have
$$S_{Qmin}=\frac{2 \sqrt{3}}{3}-2\rho+2\sqrt{(\frac{\sqrt{3}}{3}+2\rho)^2+1}$$
$$S_{Pmax}=\frac{2 \sqrt{3}}{3}+\rho+2\sqrt{(\frac{\sqrt{3}}{3}-\rho)^2+1}$$
we get
$$S_{Qmin}-S_{Pmax}= -3\rho+2\sqrt{(\frac{\sqrt{3}}{3}+2\rho)^2+1}-2\sqrt{(\frac{\sqrt{3}}{3}-\rho)^2+1} $$
We have to study the last function in the interval $0 \leq \rho \leq \frac{\sqrt{3}}{3}$ (note that higher values of $\rho$ are not allowed, because the point $Q$ would be external to the triangle $ABC$). As expected, its value for $\rho=0$ is zero (it is the case where both $P$ and $Q$ coincide with the origin $O$), whereas its value for $\rho=\frac{\sqrt{3}}{3}$ is $2-\sqrt{3}$ (it is the case where $P$ is the central point of the side $AB$, and $Q$ coincides with the point $C$). To check whether the function is positive in this range (which would prove that $S_{Qmin}\geq S_{Pmax}$), we can solve the equation
$$2\sqrt{(\frac{\sqrt{3}}{3}+2\rho)^2+1}-2\sqrt{(\frac{\sqrt{3}}{3}-\rho)^2+1}-3\rho \geq 0$$
By squaring two times, the equation above reduces to
$$\rho^3(135\rho^2-\frac{432 \sqrt{3}}{3})\leq 0$$
which has, in addition to the solution $\rho=0$, another zero in $\rho=\frac{16 \sqrt{3}}{15}=1.84752...$. The function is then positive in the interval $0<\rho<\frac{16 \sqrt{3}}{15}$, which includes our region of interest $0<\rho<\frac{\sqrt{3}}{3}$. Accordingly, here is the graph of the function (the dotted gray line shows the right boundary of the region of interest):
In conclusion, since $S_{Qmin}-S_{Pmax} \geq 0$ for all possible values of $\rho$, and because $|QA| + |QB| + |QC| \geq S_{Qmin} $ and $|PA| + |PB| + |PC| \leq S_{Pmax} $, we have
$$|PA| + |PB| + |PC| \leq |QA| + |QB| + |QC| $$
Just a final notation. Although this solution refers to the geometrical problem underlying the OP, the same procedure can also prove the algebric inequality reported at the beginning of the OP, where the radicals represents $S_Q$ and $S_P$ in a particular case. In fact, that inequality corresponds to a specific case of the general geometrical problem described above, where we assume that the polar coordinates of the points $P$ and $Q$ have opposite angles $\alpha$ and $\pi+\alpha$ (this is the consequence of assuming both cartesian coordinates of $Q$ two times larger than those of $P$ and with opposite sign, i.e., $a=-2c$ and $b=-2d$). In this case, $P$ has coordinates ($\rho \cos \alpha$, $\rho \sin \alpha$), and $Q$ has coordinates ($2\rho \cos (\pi+\alpha)$, $2\rho \sin (\pi+\alpha$)). Making the appropriate changes from cartesian to polar coordinates, we can then use the inequalities $S_Q \geq S_{Qmin}$ and $S_P \leq S_{Pmax}$ to achieve the proof. In doing this, we also have to take into account the interval over which $\rho$, and therefore also $a$ and $b$, are allowed.
