We need to prove that $4x + y \geq 4z$ given the conditions $y \geq 2z \geq 4x$ and
$$2(x^3 + y^3 + z^3) + 15(xy^2 + yz^2 + zx^2) \geq 16(x^2y + y^2z + z^2x) + 2xyz.$$

To do this, we will assume the contrary, i.e., $4x + y < 4z$, and show that it leads to a contradiction.

1. **Assume $4x + y < 4z$:**
$$4x + y < 4z$$
Let $z = 2x + b$ and $y = 4x + c$. Given $y \geq 2z \geq 4x$, we have:
$$c \geq 2b \geq 0$$
From $4x + y < 4z$, we get:
$$4x + (4x + c) < 4(2x + b) \implies 8x + c < 8x + 4b \implies c < 4b$$
Combining these, we have:
$$4b > c \geq 2b > 0 \implies 4 > k = \frac{c}{b} \geq 2$$

2. **Case 1: $k > 2$:**
Replace $(x, y, z)$ with $(x, 4x + c, 2x + b)$ in the given inequality:
$$2(x^3 + (4x + c)^3 + (2x + b)^3) + 15(x(4x + c)^2 + (4x + c)(2x + b)^2 + (2x + b)x^2) \geq 16(x^2(4x + c) + (4x + c)^2(2x + b) + (2x + b)^2x) + 2x(4x + c)(2x + b)$$
Rearrange it as a polynomial in $x$:
$$49bx^2 - (56b^2 + 7c^2 - 70bc)x - (2b^3 + 2c^3 + 15b^2c - 16bc^2) > 0$$
To prove this, we need to show that the discriminant $\Delta$ of this quadratic polynomial is negative:
$$\Delta = (56b^2 + 7c^2 - 70bc)^2 + 4 \cdot 49b \cdot (2b^3 + 2c^3 + 15b^2c - 16bc^2)$$
Simplify and check:
$$\frac{\Delta}{49} = c^4 - 12c^3b + 52c^2b^2 - 100cb^3 + 72b^4 < 0$$
Let $k = \frac{c}{b}$, then:
$$P(k) = k^4 - 12k^3 + 52k^2 - 100k + 72$$
Factorize:
$$P(k) = (k - 2)(k^3 - 10k^2 + 32k - 36)$$
Analyze $Q(k) = k^3 - 10k^2 + 32k - 36$:
$$Q'(k) = 3k^2 - 20k + 32 = (3k - 8)(k - 4)$$
$Q(k)$ is increasing on $(2, \frac{8}{3})$ and decreasing on $(\frac{8}{3}, 4)$. Evaluate at $k = \frac{8}{3}$:
$$Q\left(\frac{8}{3}\right) < 0$$
Thus, $P(k) < 0$ for $2 < k < 4$.

3. **Case 2: $k = 2$:**
This implies $c = 2b$. Substitute into the polynomial:
$$49bx^2 + 56b^2x + 16b^3 = b(7x + 4b)^2 > 0$$
If $x \neq \frac{-4b}{7}$, the inequality holds. If $x = \frac{-4b}{7}$:
$$y = 4x + 2b = \frac{-2b}{7}, \quad z = 2x + b = \frac{-b}{7}$$
This leads to $(x, y, z) = (-4t, -2t, -t)$ for $t > 0$. However, this does not satisfy $4x + y \geq 4z$.

Thus, the assumption $4x + y < 4z$ leads to a contradiction. Therefore, we conclude that:
$$4x + y \geq 4z$$
$\blacksquare$