, 2015

Consider the first two brackets and observe that
$$(x + y + z)(4x + y + 2z) = (2x + y)^2 + 3z(2x + y) + 2z^2 + xy.$$
Therefore, we can write the inequality in the form
$$\left( \frac{(2x + y)^2 + 3z(2x + y) + 2z^2}{xy} + 1 \right) \cdot \frac{2x + y + 8z}{z} \geq \frac{375}{2}.$$
Now fix $z$ and $2x + y$ and move $2x$ and $y$ closer to each other. Then we see that $xy$ increases during this movement and attains its maximum when $2x = y$.
Therefore the inequality follows from the inequality obtained by the substitution of $y = 2x$ into initial inequality, i.e.
$$(3x + z)(6x + 2z)(4x + 8z) \geq 375x^2z.$$

Letting $t = x/z$, we can rewrite the inequality in the form
$$8(3t + 1)^2(t + 2) - 375t^2 \geq 0,$$

which can be easily checked by means of derivatives. One finds the minimum to be attained for $t = 4/3$. (So the minimum in the initial inequality holds for $(x, y, z) = (4, 8, 3).$ $\square$