It is well-known that in an equilateral triangle the sum of the distances from an interior point $P$ to its sides equals the altitude of the triangle, as can be easily proven. On account of the preceding, we have to prove that if $x + y + z = 1$ then it holds that
$$x^2 + y^2 + z^2 \geq x^3 + y^3 + z^3 + 6xyz$$
To do it, we begin observing that when $x + y + z = 1$, then
$$xy + yz + zx \geq 9xyz$$
Indeed, applying AM-GM inequality, we get
$$xy + yz + zx = (xy + yz + zx)(x + y + z) \geq 3\sqrt[3]{(xy)(yz)(zx)} \cdot 3\sqrt[3]{xyz} = 9xyz$$
or
$$xy + yz + zx - 3xyz \geq 6xyz$$
On account of the constrain and the preceding inequality, we obtain
$$\begin{aligned} xy + yz + zx - 3xyz &= xy(1 - z) + yz(1 - x) + zx(1 - y) \\ &= xy(x + y) + yz(y + z) + zx(z + x) \\ &\geq 6xyz \end{aligned}$$
Adding $1$ to both terms of the last inequality and reordering terms, yields
$$(x + y + z)^2 + xy(x + y) + yz(y + z) + zx(z + x) - 6xyz \geq 1$$
or equivalently,
$$x^2 + y^2 + z^2 + 2xy(1-z) + 2yz(1-x) + 2zx(1-y) + xy(x+y) + yz(y+z) + zx(z+x) \geq 1,$$
and
$$x^2 + y^2 + z^2 + 3xy(x + y) + 3yz(y + z) + 3zx(z + x) \geq 1 = (x + y + z)^3$$
from which
$$x^2 + y^2 + z^2 \geq x^3 + y^3 + z^3 + 6xyz$$
follows. Equality holds when $x = y = z = 1/3$. That is, when $P$ is the centroid of the triangle, and we are done.