In their book Morse index of solutions of nonlinear elliptic equations, Damascelli and Pacella state and prove the following form of Hopf's lemma:
Let $B = B(y, r_1)$ be a ball, $x_0 \in \partial B$ and $v \in H^1(B) \cap C^0(B \cup \{x_0\}$ be a weak solution of $-\Delta v + c(x)v \geq 0$ in $B$, with $c \in L^\infty(B)$. Assume further that $v > 0$ in $B$ and $v(x_0) = 0$. Then $\frac{\partial v}{\partial s}(x_0) > 0$ for every inner direction $s$, in the sense that $$ \liminf_{t \to 0^+} \frac{v(x_0 + ts) - v(x_0)}{t} > 0 $$ whenever $(y - x_0) \cdot s > 0$.
The idea of the proof is to define $z(x) =e^{-\alpha r_1^2} - e^{-\alpha |x-y|^2}$ in $A = B(y, r_1) \setminus B(y, r_1/2)$, where $\alpha$ is a parameter that will later be picked big enough, and, using the continuity of the trace, apply the weak maximum principle for $w = v + \varepsilon z$ in $A$. Note that $$ \Delta z = - e^{-\alpha|x - y|^2}(4\alpha^2|x - y|^2 - 2 \alpha N) $$ Now, in a part of the proof the following chain of inequalities is presented: $$ - \Delta z + c(x) z = e^{- \alpha |x - y|^2} \left(4 \alpha^2 |x - y|^2 - 2 \alpha N + c(x) \frac{e^{-\alpha r_1^2}}{e^{- \alpha|x - y|^2}} - c(x) \right)\\ \geq e^{- \alpha |x - y|^2} \left(4 \alpha^2 \frac{r_1^2}{4} - 2 \alpha N - \|c\|_\infty \right) \\ (*) \quad \geq e^{r_1^2/4}\left(4 \alpha^2 \left(\frac{r_1^2}{4} \right)^2 - 2 \alpha N - \|c\|_\infty \right) \\ > 0 $$
I will be the most grateful for an explanation of the inequality marked with an $(*)$, namely from the second to the third line.
Thanks in advance.