Deriving $\|Du\|_{L^\infty(U)}\le C(\|Du\|_{L^\infty(\partial U)}+\|u\|_{L^\infty(\partial U)})$ with $u$ being the smooth solution to an elliptic PDE 

Let $u$ be a smooth solution of the uniformly elliptic equation $Lu=-\sum_{i,j=1}^n a^{ij}(x)u_{x_i x_j}$ in $U$. Assume that the coefficients have bounded derivatives.
Set $v:=|Du|^2+\lambda u^2$ and show that $$Lv \le 0 \quad \text{in }U$$ if $\lambda$ is large enough. Deduce $$\|Du\|_{L^\infty(U)}\le C(\|Du\|_{L^\infty(\partial U)}+\|u\|_{L^\infty(\partial U)}).$$

This is PDE Evans, 2nd edition: Chapter 6, Exercise 8.
I believe I was able to show $Lv \le 0$ for large enough $\lambda$ already. Specifically, given $v$, we obtain $$v_{x_i x_j} = 2[(Du_{x_j}\cdot Du_{x_i}+Du \cdot Du_{x_i x_j})+(\lambda u_{x_j}u_{x_i}+\lambda uu_{x_i x_j})],$$ which means, for large enough $\lambda$, $$Lv=-2\sum_{i,j=1}^n a^{ij}(x) Du_{x_j}\cdot Du_{x_i}-2\lambda\sum_{i,j=1}^n a^{ij}(x)  u_{x_j} u_{x_i} \le 0.$$
But how can I get started on deriving the estimate $$\|Du\|_{L^\infty(U)}\le C(\|Du\|_{L^\infty(\partial U)}+\|u\|_{L^\infty(\partial U)})$$
Any hints on this part would be helpful.
 A: I believe this can indeed be handled by the maximum principle.  Since $Lu,Lv\leq0$, it tells us that
$$\max_{\partial U}u=\max_{\overline{U}}u\quad\text{and}\quad\max_{\partial U}v=\max_{\overline{U}}v.$$
It helps considerably to rewrite this in terms of the infinity norm and the definition of $v$, which reads
$$\left\|u^2\right\|_{L^\infty(\partial U)}=\left\|u^2\right\|_{L^\infty(U)}\quad\text{and}\quad\left\||Du|^2+\lambda u^2\right\|_{L^\infty(\partial U)}=\left\||Du|^2+\lambda u^2\right\|_{L^\infty(U)}.$$
Applying the triangle inequality to both sides of the right equation gives
$$\left\||Du|^2\right\|_{L^\infty(\partial U)}+\left\|\lambda u^2\right\|_{L^\infty(\partial U)}\geq\left\||Du|^2\right\|_{L^\infty(U)}-\left\|\lambda u^2\right\|_{L^\infty(U)}$$
So we can then add $\lambda$ times the left equation to this, which gives
$$\left\||Du|^2\right\|_{L^\infty(\partial U)}+2\left\|\lambda u^2\right\|_{L^\infty(\partial U)}\geq\left\||Du|^2\right\|_{L^\infty(U)}$$
We can then add the appropriate terms to the left side (which are all positive, of course) to allow us to complete the square.  Taking square roots then gives exactly the desired estimate.
