[proofplan] The strategy is to construct an exponential barrier function $v$ on an annular region inside the interior sphere. The parameter in the exponential is tuned so that $Lv \le 0$ on the annulus, enabling a comparison with $u$. Because $u < u(x_0)$ throughout $U$, we obtain $u + \varepsilon v \le 0$ on the boundary of the annulus for small $\varepsilon > 0$, and the weak maximum principle extends this to the interior. Differentiating at $x_0$ — where both $u - u(x_0)$ and $v$ vanish — yields a strictly positive lower bound on $\frac{\partial u}{\partial \nu}(x_0)$. [/proofplan] [step:Set up the annular region and reduce to the case $u(x_0) = 0$] Let $B = B(y, r)$ be the open ball guaranteed by the interior sphere condition, so that $B \subset U$ and $x_0 \in \partial B$. The center $y \in U$ and radius $r > 0$ are fixed for the remainder of the proof. Define the annular region \begin{align*} A := B(y, r) \setminus \overline{B(y, r/2)}. \end{align*} Since $u \in C^1(\overline{U})$ and $u(x_0) > u(x)$ for every $x \in U$, the restriction of $u$ to the compact set $\overline{B(y, r/2)} \subset U$ attains its maximum, which is strictly less than $u(x_0)$. Hence there exists a constant $m > 0$ such that \begin{align*} u(x) \le u(x_0) - m \quad \text{for all } x \in \overline{B(y, r/2)}. \end{align*} We now reduce to the case $u(x_0) = 0$. Set $\tilde{u} := u - u(x_0)$. If $c \equiv 0$, the zeroth-order term vanishes and $L\tilde{u} = Lu \le 0$. If $c \not\equiv 0$, the sign conditions $c \ge 0$ and $u(x_0) \ge 0$ give \begin{align*} L\tilde{u} = Lu - c(x)\,u(x_0) \le 0 - c(x)\,u(x_0) \le 0. \end{align*} In either case $L\tilde{u} \le 0$ in $U$, $\tilde{u}(x_0) = 0$, and $\tilde{u} < 0$ in $U$. Replacing $u$ by $\tilde{u}$, we assume henceforth that $u(x_0) = 0$ and $u < 0$ in $U$. Under this normalization, the gap estimate becomes $u \le -m$ on $\overline{B(y, r/2)}$. [/step] [step:Construct the exponential barrier function and verify $Lv \le 0$] Define the auxiliary function $v \colon \mathbb{R}^n \to \mathbb{R}$ by \begin{align*} v(x) := e^{-\alpha |x - y|^2} - e^{-\alpha r^2}, \end{align*} where $\alpha > 0$ is a parameter to be determined. Note the key boundary values: $v(x_0) = 0$ (since $|x_0 - y| = r$) and $v > 0$ whenever $|x - y| < r$. In particular, $v > 0$ on $\partial B(y, r/2)$ and throughout the annulus $A$. We compute the derivatives of $v$. Set $\rho(x) := |x - y|^2$ and $g(x) := e^{-\alpha \rho(x)}$. Since $v$ differs from $g$ by the constant $e^{-\alpha r^2}$, all derivatives of $v$ coincide with those of $g$: \begin{align*} \partial_{x_i} v &= -2\alpha\,(x_i - y_i)\,e^{-\alpha \rho}, \\ \partial_{x_i}\partial_{x_j} v &= \bigl(4\alpha^2(x_i - y_i)(x_j - y_j) - 2\alpha\,\delta_{ij}\bigr)\,e^{-\alpha \rho}. \end{align*} Applying the operator $L$ to $v$ on $A$: \begin{align*} Lv &= -\sum_{i,j=1}^{n} a_{ij}\,\partial_{x_i}\partial_{x_j} v + \sum_{i=1}^{n} b_i\,\partial_{x_i} v + c\,v \\ &= e^{-\alpha \rho}\biggl(-4\alpha^2 \sum_{i,j} a_{ij}(x_i - y_i)(x_j - y_j) + 2\alpha \sum_{i,j} a_{ij}\,\delta_{ij} \\ &\qquad\qquad - 2\alpha \sum_{i} b_i(x_i - y_i)\biggr) + c\,v. \end{align*} We estimate each group of terms on the annulus $A$, where $r/2 \le |x - y| \le r$. **Leading quadratic term.** By uniform ellipticity with $\xi = x - y$: \begin{align*} \sum_{i,j} a_{ij}(x)(x_i - y_i)(x_j - y_j) \ge \theta\,|x - y|^2 \ge \theta\,\frac{r^2}{4}. \end{align*} This contributes $-4\alpha^2 \cdot \theta r^2/4 = -\alpha^2\,\theta\,r^2$ after multiplication by $e^{-\alpha\rho}$. **First-order and trace terms.** Set $M := \max\bigl(\sup_U \|a_{ij}\|, \sup_U |b_i|\bigr)$. On $A$ we have $|x - y| \le r$ and $|\delta_{ij}| \le 1$, so \begin{align*} \biggl|2\alpha \sum_{i,j} a_{ij}\,\delta_{ij} - 2\alpha \sum_i b_i(x_i - y_i)\biggr| \le 2\alpha\,(n^2 M + n M r) =: C_1\,\alpha, \end{align*} where $C_1 := 2n M(n + r)$ is independent of $\alpha$. **Zeroth-order term.** Since $c$ is bounded and $|v| \le 1$, we have $|c\,v| \le C_2$ for a constant $C_2 := \sup_U |c|$ independent of $\alpha$. Combining these estimates: \begin{align*} Lv \le e^{-\alpha \rho}\bigl(-\alpha^2\,\theta\,r^2 + C_1\,\alpha + C_2\bigr) \quad \text{on } A. \end{align*} The expression in parentheses is a quadratic in $\alpha$ with negative leading coefficient $-\theta\,r^2 < 0$. For all sufficiently large $\alpha$, this quadratic is negative. We fix such an $\alpha$ once and for all, obtaining \begin{align*} Lv \le 0 \quad \text{in } A. \end{align*} [/step] [step:Comparison argument on the annulus via the weak maximum principle] We compare $u$ and $\varepsilon\,v$ on the annulus $A$ for a suitably small $\varepsilon > 0$. Define $w := u + \varepsilon\,v$ on $\overline{A}$. We verify $w \le 0$ on each component of $\partial A$. **Inner sphere $\partial B(y, r/2)$.** On this set, $u \le -m$ (by the gap estimate) and $v \le \sup_A v < \infty$. Choosing \begin{align*} 0 < \varepsilon \le \frac{m}{\sup_A v}, \end{align*} we obtain $\varepsilon\,v \le m$ on $\partial B(y, r/2)$, hence \begin{align*} w = u + \varepsilon\,v \le -m + m = 0 \quad \text{on } \partial B(y, r/2). \end{align*} **Outer sphere $\partial B(y, r)$.** Here $|x - y| = r$, so $v(x) = e^{-\alpha r^2} - e^{-\alpha r^2} = 0$. Since $u \le 0$ in $\overline{U}$: \begin{align*} w = u + \varepsilon \cdot 0 = u \le 0 \quad \text{on } \partial B(y, r). \end{align*} Therefore $w \le 0$ on $\partial A$. On the interior of $A$, the linearity of $L$ gives \begin{align*} Lw = Lu + \varepsilon\,Lv \le 0 + \varepsilon \cdot 0 = 0, \end{align*} using $Lu \le 0$ (hypothesis) and $Lv \le 0$ (from the previous step). Since $c \ge 0$ and $w \le 0$ on $\partial A$, the weak maximum principle for the operator $L$ yields \begin{align*} w \le 0 \quad \text{in } A, \end{align*} that is, \begin{align*} u(x) \le -\varepsilon\,v(x) \quad \text{for all } x \in A. \end{align*} [/step] [step:Extract the strictly positive normal derivative at $x_0$] The outward unit normal to $\partial B$ at $x_0$ is $\nu = (x_0 - y)/r$. Because $B$ is tangent to $\partial U$ at $x_0$ from the interior, this coincides with the outward unit normal to $\partial U$ at $x_0$. For small $t > 0$, the point $x_0 - t\nu$ lies on the segment from $x_0$ toward $y$, which enters $A$ (since $|x_0 - t\nu - y| = r - t < r$ and, for $t$ small enough, $|x_0 - t\nu - y| = r - t > r/2$). Applying the comparison $u \le -\varepsilon v$ at $x_0 - t\nu$ and using $u(x_0) = v(x_0) = 0$: \begin{align*} \frac{u(x_0) - u(x_0 - t\nu)}{t} = \frac{-u(x_0 - t\nu)}{t} \ge \frac{\varepsilon\,v(x_0 - t\nu)}{t} = \varepsilon\,\frac{v(x_0 - t\nu) - v(x_0)}{t}. \end{align*} The left-hand side converges to $\frac{\partial u}{\partial \nu}(x_0)$ as $t \to 0^+$ (since $u \in C^1(\overline{U})$). The right-hand side converges to $\varepsilon\,\frac{\partial v}{\partial \nu}(x_0)$ evaluated from the interior. We must be careful with the sign: $\frac{\partial v}{\partial \nu}(x_0)$ is the rate of change of $v$ in the direction of increasing $|x - y|$, which is the direction of decreasing $v$. However, in the difference quotient above, we are approaching $x_0$ from the interior along $-\nu$, so the quotient $\frac{v(x_0 - t\nu) - v(x_0)}{t}$ equals \begin{align*} \frac{v(x_0 - t\nu) - 0}{t} > 0, \end{align*} since $v > 0$ in $A$. We compute this limit directly. Along the ray $x_0 - t\nu$: \begin{align*} |x_0 - t\nu - y|^2 = |x_0 - y - t\nu|^2 = r^2 - 2t\,r + t^2, \end{align*} using $\nu = (x_0 - y)/r$ so that $(x_0 - y) \cdot \nu = r$. Therefore \begin{align*} v(x_0 - t\nu) &= e^{-\alpha(r^2 - 2tr + t^2)} - e^{-\alpha r^2} \\ &= e^{-\alpha r^2}\bigl(e^{\alpha(2tr - t^2)} - 1\bigr). \end{align*} Dividing by $t$ and taking $t \to 0^+$: \begin{align*} \lim_{t \to 0^+} \frac{v(x_0 - t\nu)}{t} = e^{-\alpha r^2} \cdot \lim_{t \to 0^+} \frac{e^{\alpha(2tr - t^2)} - 1}{t} = e^{-\alpha r^2} \cdot 2\alpha r. \end{align*} The limit uses $\lim_{t \to 0} (e^{f(t)} - 1)/t = f'(0)$ with $f(t) = \alpha(2tr - t^2)$ and $f'(0) = 2\alpha r$. Combining with the inequality: \begin{align*} \frac{\partial u}{\partial \nu}(x_0) \ge \varepsilon \cdot 2\alpha\,r\,e^{-\alpha r^2} > 0. \end{align*} Since $\varepsilon > 0$, $\alpha > 0$, $r > 0$, and $e^{-\alpha r^2} > 0$, the right-hand side is strictly positive. [/step]