As know the 3 circle concur on a point P. Let a homothety with center P and ratio 2, this send A' on A'', B' on B'' and C' on C'' s.t. A'', B'' and C'' stay respectively on the circumcircle of AYZ, BZX, CXY and we have that X stay on B''C'', Y on C''A'' and Z on A''B''. We have also $\angle APA'' = \angle AZA'' = \angle BPB'' = \angle BXB'' = \angle CPC''$, then we can obtain A''B''C'' from ABC with a spiral dilatation with center P, angle $\angle APA''$ and ratio $\frac{PA''}{PA}$, but $PA'' \ge PA$ so we have $4(A'B'C') = (A''B''C'') \ge (ABC)$. If we have equality we have $PA'' = PA$ and so $A \equiv A''$, and so also $B \equiv B''$, $C \equiv C''$, so AA', BB', CC' concur on P. If AA', BB', CC' concur then the perpendicolar from A to YZ, the perpendicolar from B to ZX and from C to XY are concurrent on the isogonal conjugate of the point of concurrence and so ABC and XYZ are ortholigic triangles, so XYZ is the pedal triangle of P so the equality follows.

i have a different solution, although very ugly... suppose wlog that angles $A,B$ are acute angle... set a coordinate system with $A = ( - 2a,0), B = (2b,0), C = (0,2c)$, with $a,b,c > 0$, and let $X = (2by, 2c(1 - y)), Y = ( - 2az, 2c(1 - z)), Z = ( - 2a(1 - x) + 2bx,0)$... the coordinates of the circumcenters are $\left( - a(2 - x) + bx,\dfrac{a^2(1 - x - z) - abx + c^2(1 - z)}{c}\right), \left( - a(1 - x) + b(1 + x), \dfrac{ - ab(1 - x) + b^2(x - y) + c^2(1 - y)}{c}\right)$, $\left( \dfrac{-a^2z+b^2y+c^2(y-z)}{a+b}, \dfrac{-a^2bz-ab^2y+ac^2(2-y)+bc^2(2-z)}{c(a+b)} \right)$ after a couple of hours of algebraic manipulation we arrive that the inequality is equivalent to $(a^2(1 - x - z) + b^2(y - x) + c^2(y - z) + ab(1 - 2x))^2\geq 0$, which is true...

by the miquel theorem, the circumcircles of the triangles AYZ, BZX, CXY are a pencil of circles, call M the common point. by symetry it's easy to see that the angle A'B'C' is a half of the arc ZX (circumcircle of BXZ), so it is equal to the angle ABC. similarly angle B'C'A'= angle BCA, then the triangles ABC and A'B'C' are similar. now the problem is equivalent to prove that 2.B'C' is bigger or equal to BC (and similarly with the other sides). again by symetry, the angle MB'C' is a half of the arc MX (circumcircle of BXZ) and then it is equal to the angle MBC. similarly the angle MC'B'= angle MCB. that implies the triangles MB'C' and MBC are similar and B'C'/BC=B'M/BM. but 2.B'M is bigger or equal to BM because 2.B'M is a diameter and BM a chord (circumcircle of BXZ), and the result follows. the equality holds if and only if BM, CM, AM are diameters of respective circumcircles, and by thales it's easy to prove that the triangles A'B'C' and ABC are homothetic, so AA', BB', CC' are a pencil of straight lines.