Did you mean that the length of OQ is twice the length of PO?

And $O\in PQ$.

Let $J,K,L$ be the midpoints of $BC,CA,AB$ respectively. Then it is clear that $AQ=2JP$, and by Ptolemy's Theorem, $AK\cdot LP+AL\cdot KP\ge KL\cdot AP$. But $\triangle AKL$ is equilateral, so $LP+KP\ge AP$. Summing up cyclically we have the desired result.

Also because $AP$, $KP$ and $LP$ are sides-lengths of triangle.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -21.799688221040096, xmax = 25.38028625757998, ymin = -11.219412740040317, ymax = 12.023662922221074; /* image dimensions */ pen qqccqq = rgb(0,0.8,0); /* draw figures */ draw((-8.28,7.41)--(-12.518883462392374,0.1373205080756903), linewidth(1)); draw((-12.518883462392374,0.1373205080756903)--(-4.101116537607631,0.10267949192430858), linewidth(1)); draw((-4.101116537607631,0.10267949192430858)--(-8.28,7.41), linewidth(1)); draw((-7.912294734194974,4.381525218736299)--(-10.08,1.97), linewidth(1)); draw((-10.08,1.97)--(-6.907705265805028,1.2984747812636992), linewidth(1)); draw((-6.907705265805028,1.2984747812636992)--(-7.912294734194974,4.381525218736299), linewidth(1)); draw((-7.912294734194974,4.381525218736299)--(-12.518883462392374,0.1373205080756903), linewidth(1) + qqccqq); draw((-12.518883462392374,0.1373205080756903)--(-7.41,2.84), linewidth(1) + qqccqq); draw((-12.518883462392374,0.1373205080756903)--(-6.907705265805028,1.2984747812636992), linewidth(1) + qqccqq); draw((-7.912294734194974,4.381525218736299)--(-2.3011165376076263,5.542679491924309), linewidth(1) + qqccqq); draw((-2.3011165376076263,5.542679491924309)--(-7.41,2.84), linewidth(1) + qqccqq); draw((-2.3011165376076263,5.542679491924309)--(-6.907705265805028,1.2984747812636992), linewidth(1) + qqccqq); /* dots and labels */ dot((-8.28,7.41),linewidth(1pt) + dotstyle); label("$B$", (-8.188510462718474,7.452592730520854), NE * labelscalefactor); dot((-8.3,2.55),linewidth(1pt) + dotstyle); label("$O$", (-8.208917026074278,2.595830651839369), NE * labelscalefactor); dot((-12.518883462392374,0.1373205080756903),linewidth(1pt) + dotstyle); label("$A$", (-12.433075640725816,0.18785617585443112), NE * labelscalefactor); dot((-4.101116537607631,0.10267949192430858),linewidth(1pt) + dotstyle); label("$C$", (-4.02557153813435,0.147043049142822), NE * labelscalefactor); dot((-10.08,1.97),linewidth(1pt) + dotstyle); label("$Q$", (-10.004694601385077,2.004040314521037), NE * labelscalefactor); dot((-7.41,2.84),linewidth(1pt) + dotstyle); label("$P$", (-7.331434801774684,2.881522538820633), NE * labelscalefactor); dot((-6.907705265805028,1.2984747812636992),linewidth(1pt) + dotstyle); label("$Q'$", (-6.821270717879571,1.3306237237794864), NE * labelscalefactor); dot((-7.912294734194974,4.381525218736299),linewidth(1pt) + dotstyle); label("$Q''$", (-7.821192322313993,4.43242135386178), NE * labelscalefactor); dot((-8.996147367097489,3.175762609368151),linewidth(1pt) + dotstyle); label("$P'$", (-8.923146743527437,3.2080275525135055), NE * labelscalefactor); dot((-8.493852632902513,1.6342373906318483),linewidth(1pt) + dotstyle); label("$P''$", (-8.412982659632323,1.677535300828164), NE * labelscalefactor); dot((-2.3011165376076263,5.542679491924309),linewidth(1pt) + dotstyle); label("$X$", (-2.209387399467747,5.575188901786834), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Draw equilateral triangles $QQ'Q''$ and $PP'P''$ centered at $O$. Let $P$ be the midpoint of $AX$. Since $P$ is the midpoint of $Q'Q''$, $AQ'XQ''$ is a parallelogram. By triangle inequality, $|AQ'|+|AQ''|=|AQ'|+|Q'X|\ge |AX|=2|AP|$. Summing cyclically, $LHS=|AP|+|AP'|+|AP''|\le |AQ|+|AQ'|+|AQ''|=RHS$.