First of all, for instance, $x < y + z$, because
$$2x = b + c < 2a + b + c = (c + a) + (a + b) = 2y + 2z.$$
Thus, $x, y, z$ are the side lengths of a triangle. Next, $x + y + z = a + b + c = 2s$, in the usual notation. Hence, $\triangle ABC$ and $\triangle XYZ$ have the same perimeter. Now, by Heron's formula, the area of $\triangle XYZ$ is given by $(XYZ) = \sqrt{s(s-x)(s-y)(s-z)}$. But,
$$s-x = \frac{2s-(b+c)}{2} = \frac{a}{2}, \quad s-y = \frac{2s-(c+a)}{2} = \frac{b}{2}, \quad s-z = \frac{2s-(a+b)}{2} = \frac{c}{2}$$
and so
$$(s-x)(s-y)(s-z) = \frac{abc}{8}.$$
Thus, $(XYZ)^2 = abc/s$, and so, since $(ABC)^2 = s(s-a)(s-b)(s-c)$, we must show that
$$abc \geq 8(s-a)(s-b)(s-c),$$
with equality iff $a = b = c$. Now
$$2\sqrt{(s-a)(s-b)} \leq s-a+s-b=c,$$
with equality iff $a = b$. Hence,
$$8(s-a)(s-b)(s-c) = (2\sqrt{(s-a)(s-b)}) (2\sqrt{(s-b)(s-c)}) (2\sqrt{(s-c)(s-a)}) \leq cab,$$
with equality iff $a = b = c$.