Semilinear Poisson PDE - proving a (hopefully) simple inequality This is from page 557 of PDE Evans, 2nd edition. My question is at the bottom of this post, but for now, here is some context for my question:

LEMMA 2 (Boundary estimates). Let $u \in C^2(\bar{U})$ satisfy $\text{(2)}$, $\text{(3)}$. Then for each point $x^0 \in \partial U \cap \{x_n > 0\}$, either $$u_{x_n}(x^0) < 0 \tag{7}$$ or else $$u_{x_n}(x^0)=0, \, u_{x_n x_n}(x^0)>0. \tag{8}$$ In either case, $u$ is strictly decreasing as a function of $x_n$ near $x^0$.

For reference, here are $\text{(2)}$ and $\text{(3)}$ from the book:

In this subsection we take $U = B^0(0,1)$ to be the open unit ball in $\mathbb{R}^n$ and investigate this boundary-value problem for a semilinear Poisson PDE: $$\begin{cases}-\Delta u = f(u) & \text{in }U \\ \quad \, \, \,  u=0 & \text{on }\partial U. \end{cases} \tag{2}$$  We are interested in positive solutions: $$u > 0 \quad \text{in }U \tag{3}$$ and will assume $f : \mathbb{R} \to \mathbb{R}$ is Lipschitz continuous but otherwise arbitrary.

Now, here is a fragment of the proof to Lemma 2:

Proof. 1. Fix any point $x^0 \in \partial U \cap\{x_n > 0\}$ and let $\nu = \mathbf{\nu}(x^0)=\nu_1,\ldots,\nu_n)$ denote the outer unit normal to $\partial U$ at $x^0$. Note $\nu_n > 0$.
  2. We first claim $$u_{x_n}(x^0)<0,$$ provided $$f(0) \ge 0. \tag{9}$$ Indeed \begin{align}0=-\Delta u -f(u) &=-\Delta u-f(u)+f(0)-f(0) \\ &\le -\Delta u + cu, \end{align} for $c(x) := -\int_0^1 f'(su(x)) \, ds$.

My question:
How does one obtain this inequality $0 \le -\Delta u + cu$? Here is what I did so far (in attempting to add more details):
\begin{align}
0 &= -\Delta u - f(u) \\
&= -\Delta u - f(0) - (f(u)-f(0)) \\
&= -\Delta u - f(0) + c \\
&\le -\Delta u - f(0) + cu \qquad (u \ge 0  \text{ in $U$ by (3)}) \\
&\le -\Delta u + 0 + cu \\
&= -\Delta u + cu
\end{align}
I'm not sure about my first inequality in which I tried to justify using $u \ge 0$. That probably does not conclude the next line.

Actually, $f(u(x))-f(0)$ is equal to $-cu(x)$, not to $-c$. Indeed,
$$-cu(x) = u(x)\int_0^1 f'(su(x)) \, ds = \int_0^1 \frac{d}{ds}
\left(f(su(x))\right) \, ds = f(u(x))-f(0)$$
where the second equality is the chain rule. 
(Keep in mind that in general, $f'(2x)$ means $f'$ evaluated at $2x$, not the derivative of $f(2x)$.)
So, 
$$-\Delta u-f(u)+f(0)-f(0) \le -\Delta u-f(u)+f(0) = -\Delta u+cu$$

