PDE Evans, 2nd edition: Chapter 6, Exercise 11
Assume $u \in H^1(U)$ is a bounded weak solution of $$-\sum_{i,j=1}^n (a^{ij}u_{x_i})_{x_j} = 0 \quad \text{in }U.$$ Let $\phi : \mathbb{R} \to \mathbb{R}$ be convex and smooth, and set $w=\phi(u)$. Show $w$ is a weak subsolution; that is, $B[w,v] \le 0$ for all $v \in H_0^1(u)$, $v \ge 0$.
Attempted proof:
Given $w=\phi(u)$, we find the derivative $w_{x_i}=\phi'(u) u_{x_i}$. Also, since $\phi$ is convex, $\phi''(u) \ge 0$. We use integration by parts to deduce that \begin{align*} B[w,v]&=\int_U \sum_{i,j=1}^n a^{ij} w_{x_i} v_{x_j} \, dx \\ &= \int_U \sum_{i,j=1}^n a^{ij} \phi'(u) u_{x_i} v_{x_j} \, dx \\ &= -\int_U \sum_{i,j=1}^n (a^{ij} \phi'(u) u_{x_i})_{x_j} v \, dx \\ &= -\int_U \sum_{i,j=1}^n a^{ij} u_{x_i} [\phi'(u)]_{x_j} v + \sum_{i,j=1}^n (a^{ij} u_{x_i})_{x_j} \phi'(u) v \, dx \\ &= -\int_U \sum_{i,j=1}^n a^{ij} \phi''(u) u_{x_i} u_{x_j} v \, dx \\ &= -\int_U \phi''(u) v \sum_{i,j=1}^n a^{ij} u_{x_i} u_{x_j} \, dx \\ &\le 0 \end{align*} for all $v \in H_0^1(U)$, $v \ge 0$.
Question: I am confident that I started correctly. However, my very last inequality is not clear. I tried to expand my penultimate step using a product-to-sum inequality (i.e. Minkowski's inequality), but then again I have a minus sign and I don't know if my terms are non-negative.
A standing assumption in exercises in that chapter is that the coefficients $a^{ij}$ satisfy an ellipticity condition. In particular, the matrix $A=(a^{ij})$ is positive definite, which implies
$$\sum_{i,j=1}^n a^{ij} u_{x_i} u_{x_j} = \nabla u^T A\nabla u\ge 0$$ Since also $v\ge 0$ and $\phi''\ge 0$, the conclusion $$-\int_U \phi''(u) v \sum_{i,j=1}^n a^{ij} u_{x_i} u_{x_j} \, dx \le 0$$ follows.