We want to show that for any point $\alpha_0 \neq \alpha_m$ the interval $J_2$ coming from any small interval $I_2$ centred on $\alpha_0$ is larger than the interval $J_1$ coming from the interval $I_1$ of the same width as $I_2$, but centred on $\alpha_m$ (see the figure). Since $\alpha_m$ is the only point with $D_f^\prime(\alpha)=0$ we know that $D_f(\alpha_0)=s \neq 0.$ To match with our figure we'll assume that $s0.$ Now choose a positive number $\epsilon$ with $\epsilon 0$ such that for all $\alpha$ with $|\alpha-\alpha_m|0$ such that for all $\alpha$ with $|\alpha-\alpha_0|(s+\epsilon)(\alpha^\prime_1-\alpha_0) +(s+\epsilon)(\alpha_0-\alpha^\prime_2)=-2\delta(s+\epsilon).\end{equation} Since $\alpha^\prime$ lies within $\delta_1$ of $\alpha_m$, we know from inequality 1 that \begin{equation}D_f(\alpha^\prime)-D_f(\alpha_m)-2\delta(s+\epsilon)>\epsilon\delta>D_f(\alpha^\prime)-D_f(\alpha_m).$$ Thus, interval $J_2$ is wider than interval $J_1$. QED.

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