Post #3142160

\[(1+\epsilon)^2\leqslant q<\left(\frac{(2)\cdot(14)\cdot\left\{2k+1+(k-1)\epsilon\right\}}{9(k+2)\cdot(2+(1-k)\epsilon^{\prime})}\right)\leqslant \left(\frac{5}{3}\right)\] \[\qquad \because(1+\epsilon)< \left(\frac{14}{9}\right)\approx q \leqslant \left(\frac{5}{3}\right)\]