The derivative of the energy with respect to $\alpha$ gives

$$\begin{equation*} \frac{d\langle E_L[\alpha]\rangle}{d\alpha} = \frac{1}{2}\alpha-\frac{1}{2\alpha^3} \end{equation*}$$

and a second derivative which is always positive (meaning that we find a minimum)

$$\begin{equation*} \frac{d^2\langle E_L[\alpha]\rangle}{d\alpha^2} = \frac{1}{2}+\frac{3}{2\alpha^4} \end{equation*}$$

The condition

$$\begin{equation*} \frac{d\langle E_L[\alpha]\rangle}{d\alpha} = 0, \end{equation*}$$

gives the optimal $\alpha=1$, as expected.