the time averaged total energy, $\bar E$, has the following $\varepsilon$ expansion in $D$ dimension: \begin{equation} \bar{E}=\varepsilon^{2-D}\frac{E_0}{2\lambda}+ \varepsilon^{4-D}E_1 \end{equation} from the above equation how can we write \begin{equation} \omega_{\rm m}^2= 1-\frac{1}{2\lambda}\frac{(D-2)E_0}{(4-D)E_1}\,. \end{equation} where $$\omega_{\rm m}^2=1-\varepsilon_{\rm m}^2,$$ and $\lambda$ is a constant factor. m denotes that angular frequency $\omega $ is minimum.
Can someone explain me how equation (2) can be written from equation (1) If you feel problem or need more information then please comment. thanks in advance. for more info see here in equation (39 and 40)
Explaining Antoine's suggestion:
Differentiate $\bar E$ with respect to $\epsilon$ to find its optimal value. You get $(2-D)\epsilon^{1-D} E_0/(2\lambda)+(4-D)\epsilon^{3-D}E_1=0$. Factor $\epsilon^{1-D}$ from that to get $(2-D)E_0/2\lambda+(4-D)\epsilon^2E_1=0$. Now solve for $\epsilon^2$ to get $\epsilon^2=-E_0/2\lambda E_1(2-D)/(4-D)$. Then $1-\epsilon^2=1+E_0/2\lambda E_1(2-D)/(4-D)$ which is ame as your final answer.