How prove this inequality $\sqrt{(2a+1)^2+(2b-\frac{\sqrt{3}}{3})^2}+\sqrt{(2a-1)^2+(2b-\frac{\sqrt{3}}{3})^2}+\cdots$ Question:

let $a,b\in R$, show that 
$$\sqrt{(2a+1)^2+(2b-\dfrac{\sqrt{3}}{3})^2}+\sqrt{(2a-1)^2+(2b-\dfrac{\sqrt{3}}{3})^2}+\sqrt{4a^2+(2b+\dfrac{2\sqrt{3}}{3})^2}\ge \sqrt{(a-1)^2+(b+\dfrac{\sqrt{3}}{3})^2}+\sqrt{(a+1)^2+(b+\dfrac{\sqrt{3}}{3})^2}+\sqrt{a^2+(b-\dfrac{2\sqrt{3}}{3})^2}$$

My idea: 
This inequality is from this problem:
let $O$ be the center of the equilateral triangle $\Delta ABC$, and for any point $P,Q$ such
$$\overrightarrow{OQ}=2
\overrightarrow{PO}$$
show that
$$|PA|+|PB|+|PC|\le |QA|+|QB|+|QC|$$

my try: let $$|AB|=|BC|=|AC|=2,O(0,0),A(1,-\dfrac{\sqrt{3}}{3}),B(-1,-\dfrac{\sqrt{3}}{3}),C(0,\dfrac{2\sqrt{3}}{3})$$
let $P(a,b),Q(c,d)$,since
$$\overrightarrow{OQ}=2
\overrightarrow{PO}\Longrightarrow a=-2c,b=-2d$$
then 
$$|PA|+|PB|+|PC|=\sqrt{(a-1)^2+(b+\dfrac{\sqrt{3}}{3})^2}+\sqrt{(a+1)^2+(b+\dfrac{\sqrt{3}}{3})^2}+\sqrt{a^2+(b-\dfrac{2\sqrt{3}}{3})^2}$$
and
$$|QA|+|QB|+|QC|= \sqrt{(c-1)^2+(d+\dfrac{\sqrt{3}}{3})^2}+\sqrt{(c+1)^2+(d+\dfrac{\sqrt{3}}{3})^2}+\sqrt{c^2+(d-\dfrac{2\sqrt{3}}{3})^2}$$
But I can't
 A: I will give a solution for the original geometric inequality.
Let $D,E,F$ be the midpoints of sides $BC,CA,AB$ respectively. Note that scaling everything by a factor of $-\frac12$ sends $\triangle ABC$ to $\triangle DEF$, and sends $Q$ to $P$. Hence $QA=2PD$, $QB=2PE$, and $QC=2PF$.
Now by Ptolemy's inequality on $APEF$, we have $PE \cdot AF + PF \cdot AE \geq PA \cdot EF$. Since $AEF$ is equilateral, this implies $PE+PF \geq PA$. Similarly,  $PF+PD \geq PB$ and  $PD+PE \geq PC$. Adding these three inequalities together gives $$QA+QB+QC = 2(PD+PE+PF) \geq PA+PB+PC$$ as desired.
Equality holds when $APEF,BPFD,CPDE$ are all concyclic, i.e. when $P$ lies on the intersection of the circumcircles of $AEF,BFD,CDE$. This happens at exactly one point, $P=O$.
A: Proof without words ( partial answer / informal proof ) .
Consider the following function:
$$
f(a,b) = \sqrt{(2a+1)^2+(2b-\dfrac{\sqrt{3}}{3})^2}
       + \sqrt{(2a-1)^2+(2b-\dfrac{\sqrt{3}}{3})^2}
       + \sqrt{4a^2+(2b+\dfrac{2\sqrt{3}}{3})^2} \\
       - \sqrt{(a-1)^2+(b+\dfrac{\sqrt{3}}{3})^2}
       - \sqrt{(a+1)^2+(b+\dfrac{\sqrt{3}}{3})^2}
       - \sqrt{a^2+(b-\dfrac{2\sqrt{3}}{3})^2}
$$

Contour lines darker if closer to zero; coordinate axes in yellow.

Picture on the left: contours of the function for $-3 < a < +3$ and $-3 < b < +3$ ;
contour levels $\in \left\{ 0.01,0.1,0.4,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15 \right\}$ ;
the higher level lines appear to be circles.

Picture on the right: contours of the function for $-0.1 < a < +0.1$ and $-0.1 < b < +0.1$ ; contour levels $\in \left\{ 0.01,0.1,0.4,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15 \right\} / 200$ ; the low level lines appear to be circles as well (or ellipses ? ).
Brute force graphical/numerical evidence thus shows that the function is positive everywhere and that the minimum is expected at $(a,b) = (0,0)$ ;
indeed $f(0,0) = 2\sqrt{1+1/3} + \sqrt{4/3} - 2\sqrt{1+1/3} - \sqrt{4/3} = 0$ .
Yes, I know this is not what people here finally want,
but I hope it helps to find a more satisfactory answer.

Geometrical meaning. As already present in the OP's question,
repeated here for convenience.
Let the origin be $O = (0,0)$ . $A = (-1,-\sqrt{3}/3)$ , $B = (+1,-\sqrt{3}/3)$ ,
$C = (0,+2\sqrt{3}/3)$ , which is an equilateral triangle with length of the edges $=2$ .
We choose a random point $\,\color{green}{Q = (a,b)}\,$
and define another point $\,\color{red}{P = (-2a,-2b)}$ .
The yellow lines are the coordinate system and $\overline{PQ}$ going through the origin $O$.

Then we have to prove that the sum of the red lengths is greater than the sum of the green lengths: 
$\overline{PA} + \overline{PB} + \overline{PC} \ge \overline{QA} + \overline{QB} + \overline{QC}$ .
Which is beyond me at the moment being.
