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 $$ S := (y+z) x + (z+x) y + (x+y) z, \quad T := \frac1{x^3} + \frac1{y^3} + \frac1{z^3}. $$$$ S := (y+z) x + (z+x) y + (x+y) z, $$ $$ T := \frac1{x^{3/2}} + \frac1{y^{3/2}} + \frac1{z^{3/2}}. $$ 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, $$ \frac1{x^3} + 2 \geq \frac3x $$$$ 2 \frac1{x^{3/2}} + 1 \geq \frac3x $$ with equality $\Leftrightarrow x = 1$, and likewise for $y$ and $z$. Therefore $$ T \geq \frac32 S - 6, $$$$ 2T \geq \frac32 S - 3, $$ and it remains to prove that $$ S \geq 6 \Rightarrow \left(\frac32 S - 6\right)^2 \geq \frac32 S $$$$ S \geq 6 \Rightarrow \frac14 \left(\frac32 S - 3\right)^2 \geq \frac32 S $$ with equality $\Leftrightarrow S = 6$. But this is clear from the factorization $$ \left(\frac32 S - 6\right)^2 - \frac32 S = \frac34 (S-6) (3S-8) $$$$ \frac14 \left(\frac32 S - 3\right)^2 - \frac32 S = \frac3{16} (S-6) (3S-2) $$ since $S \geq 6 \Rightarrow 3S-8 \geq 10 > 0$$S \geq 6 \Rightarrow 3S-2 \geq 16 > 0$. QED