I proposed the conjecture as following:
Let $ABC$ be a triangle, $P$ be arbitrary point inside of $ABC$, let $A_1B_1C_1$ be the tangential of ABC. Let A', B', C' be three be arbitrary points on $B_1C_1, C_1A_1, A_1B_1$ respectively. Let internal angle bisectors of three angles $\angle B'PC'$, $\angle C'PA'$, $\angle A'PC'$ meet $BC, CA, AB$ at $A'', B'', C''$ respectively. Then show that: $$PA'+PB'+PC' \ge 2(PA''+PB''+PC'')$$. Equality hold iff ABC be an equilateral triangle and $\angle B'PC'= \angle C'PA'=\angle A'PC' =120^0$ When $A'=A, B'=B, C'=C$ we have Barrow's inequality
Can you check in The Meterial Geogebra
