Below we prove a stronger result: the incenter $I$ of triangle $ABC$ lies inside the incircle of triangle $XYZ$. Let $d(U, VW)$ denote the distance from point $U$ to line $VW$.
Let $O$ be the incenter of $\triangle XYZ$ and let $r, r'$ and $R', R''$ denote respectively the inradius of $\triangle ABC$, the inradius of $\triangle XYZ$, and the circumradius of $\triangle XYZ$. Then $R' = 2r'$, and what we want to prove is equivalent to proving the inequality $OI \le r'$. Assume $O \ne I$ (if $O = I$ the result is obvious).
Let the incircle of $\triangle ABC$ be tangent to sides $BC, AC, AB$ at $A_1, B_1, C_1$ respectively. The lines $IA_1, IB_1, IC_1$ divide the plane into 6 acute angles, each of which contains one of the points $A_1, B_1, C_1$. We may assume that $O$ lies in the angle determined by lines $IA_1$ and $IC_1$ that contains the point $C_1$ (as in Figure 1). Let $A'$ and $C'$ be respectively the projections of $O$ onto lines $IA_1$ and $IC_1$.
Since $OX = R'$, $d(O, BC) \le R'$. Since $OA' \parallel BC$, we obtain
$$d(A', BC) = A'I + r \le R' \text{ or } A'I \le R' - r.$$
On the other hand, the incircle of $\triangle ABC$ lies inside $\triangle ABC$. Hence $d(O, AB) \ge r'$. Similarly we obtain
$$d(O, AB) = C'C_1 = r - IC' \ge r' \text{ or } IC' \le r - r'.$$
Since both $\angle A'$ and $\angle C'$ are right angles, the quadrilateral $IA'OC'$ is cyclic (as in Figure 1).
On this circle, arc $A'OC'$ $= 2\angle A'IC'$ < $180^\circ$ = arc $OC'I$, so $180^\circ \ge$ arc $IC' >$ arc $A'O$. This means $IC' > A'O$. From this we obtain
$$OI \le IA' + A'O < IA' + IC' \le (R' - r) + (r - r') = R' - r' = r' \text{ , as desired!}$$

Fig. 1

Fig. 2