Is the "cauchy-schwarz-inequality" tag a guess or a hint? . . .

At any rate, it turns out to be a good start.  Let
$$
R := \frac1{(y+z) x^4} + \frac1{(z+x) y^4} + \frac1{(x+y) z^4}.
$$
We show $xyz = 1 \Rightarrow R > 3/2$, with equality if and only if
$(x,y,z) = (1,1,1)$.  By Cauchy-Schwarz, $RS \geq T^2$, where
$$
S := (y+z) x + (z+x) y + (x+y) z,
$$ $$
T := \frac1{x^{3/2}} + \frac1{y^{3/2}} + \frac1{z^{3/2}}.
$$
Note that
$$
S = 2(yz + zx + xy) = 2\left(\frac1x + \frac1y + \frac1z\right);
$$
because $xyz = 1$; in particular, $S \geq 6$ by the AM-GM inequality,
with equality $\Leftrightarrow (x,y,z) = (1,1,1)$.
By weighted AM-GM,
$$
2 \frac1{x^{3/2}} + 1 \geq \frac3x
$$
with equality $\Leftrightarrow x = 1$, and likewise for $y$ and $z$.
Therefore
$$
2T \geq \frac32 S - 3,
$$
and it remains to prove that
$$
S \geq 6 \Rightarrow \frac14 \left(\frac32 S - 3\right)^2 \geq \frac32 S
$$
with equality $\Leftrightarrow S = 6$.
But this is clear from the factorization
$$
\frac14 \left(\frac32 S - 3\right)^2 - \frac32 S = \frac3{16} (S-6) (3S-2)
$$
since $S \geq 6 \Rightarrow 3S-2 \geq 16 > 0$.  **QED**