We wish to determine fixed $ a,b,c,d,e,f$ to satisfy the inequality for all nonnegative rational $ R$. As $ R \rightarrow \sqrt[3]{2}$ through a sequence of rational numbers, the right hand side of this inequality approaches zero. Consequently, the left hand side must vanish if we set $ R=\sqrt[3]{2}$. Hence, \[ a \cdot 2^{\frac{2}{3}} + b \cdot 2^{\frac{1}{3}} + c = 2d + e \cdot 2^{\frac{2}{3}} + f \cdot 2^{\frac{1}{3}}.\] It follows that $ a=e$, $ b=f$, $ c=2d$. On substituting back into the inequality and factoring out the common factor $ R - \sqrt[3]{2}$ from both sides, we obtain \[ \left | \frac{aR + b-d \cdot 2^{\frac{1}{3}} \left (R + 2^{\frac{1}{3}} \right )}{dR^2+aR+b} \right | < 1.\] For the last inequality to be satisfied, it suffices to let $ a,b,d$ be positive integers and make the numerator nonnegative, i.e., by letting $ a>d \cdot 2^{\frac{1}{3}}$, $ b > d \cdot 2^{\frac{2}{3}}$. A simple choice is $ d=1$, $ a=b=2$, leading to \[ \frac{2R^2+2R+2}{R^2+2R+2}.\]

Note that we must have $\frac{aR^2+bR+c}{dR^2+eR+f} < R$, so $aR^2+bR+c<dR^3+eR^2+fR$. This is $dR^3+(e-a)R^2+(f-b)R-c>0$. We can simply set $d=1$ and $a=b=2$ to make this work out; the only issue is with $R$ being negative and $dR^3+(f-b)R-c$ being too small, and that's covered.

We wish to determine fixed $ a,b,c,d,e,f$ to satisfy the inequality for all nonnegative rational $ R$. As $ R \rightarrow \sqrt[3]{2}$ through a sequence of rational numbers, the right hand side of this inequality approaches zero. Consequently, the left hand side must vanish if we set $ R=\sqrt[3]{2}$. Hence, \[ a \cdot 2^{\frac{2}{3}} + b \cdot 2^{\frac{1}{3}} + c = 2d + e \cdot 2^{\frac{2}{3}} + f \cdot 2^{\frac{1}{3}}.\] It follows that $ a=e$, $ b=f$, $ c=2d$. On substituting back into the inequality and factoring out the common factor $ R - \sqrt[3]{2}$ from both sides, we obtain \[ \left | \frac{aR + b-d \cdot 2^{\frac{1}{3}} \left (R + 2^{\frac{1}{3}} \right )}{dR^2+aR+b} \right | < 1.\] For the last inequality to be satisfied, it suffices to let $ a,b,d$ be positive integers and make the numerator nonnegative, i.e., by letting $ a>d \cdot 2^{\frac{1}{3}}$, $ b > d \cdot 2^{\frac{2}{3}}$. A simple choice is $ d=1$, $ a=b=2$, leading to \[ \frac{2R^2+2R+2}{R^2+2R+2}.\]