#Min-1

Multiplying the nominator and denominator of the summand by $x$ , we have $\displaystyle \sum_{\text{cyc}} \frac {x^5}{x^3yz+yz^4} = \sum_{\text{cyc}} \frac {x^6}{x^4yz+xyz^4}$ .

Using Titu's lemma :

$\begin{aligned} \sum_{\text{cyc}} \frac {x^6}{x^4yz+xyz^4} & \ge \frac {\left(x^3+y^3+z^3\right)^2}{x^4yz+xyz^4+xy^4z+x^4yz+xyz^4 +x^4yz} \\ & \ge \frac {\left(x^3+y^3+z^3\right)^2}{2xyz\left(x^3+y^3+z^3\right)} \\ & \ge \frac {x^3+y^3+z^3}{2xyz} \quad \quad \small \color{#3D99F6} \text{Equality occurs when }x=y=z \\ & \ge \frac 32 = \boxed{1.5} \end{aligned}$