Let us fix the directions of the blue and red segments and consider the function $$f(P)=PA’+PB’+PC’-2(PA’’+PB’’+PC’’).$$It is easy to see that it will be linear, therefore, to prove the inequality $f(X)\geqslant 0$, it is necessary to check it at the vertices A, B, C. Let us check at the vertex A (similarly at the other two). Let us prove a stronger inequality: $AB’\cdot AC’ \geqslant (AA’’)^2$. Let us put a point $X$ on $AC’$ so that $AB’\cdot AX= (AA’’)^2$. Note that the composition of the inversion with center $A$ and radius $AA’’$ and symmetry with respect to $AA’’$ maps $B’$ to $X$, then $X$ lies on the image of the tangent at point $B$ to the circle $(ABC)$. Let us show that this circle $(DAE)$, $D$ is a point on the tangent to the circle $(ABC)$ at point $C$, such that $AA'CD$ is cyclic, $E$ is a point on $AA'$, such that $DE\parallel CB$. This is easy to understand, since $AD_0E_0$ is similar to $AED$, and $AD_0A’’$ and $AA’’D$ are also similar, $D_0$ is a point on the tangent to the circle $(ABC)$ at point $B$, such that $AA'BD_0$ is cyclic, $E_0=AA' \cap BD_0$. Note (this is obvious) that circle $(DAE)$ is tangent to $CE$, then $X$ lies on the segment $AC’$. Otherwise, $PA’’$ is the external, not the internal, bisector of angle $B’PC’$. Thus, $AC’ \geqslant AX$, which gives us the solution of the problem.

Picture for the solusion: https://i.sstatic.net/p4AV8Ifg.jpg