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

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, $P=O$.
