How to prove that $\sum_{i,j,k,l=1}^n a^{ij} a^{kl} v_{x_i x_k} v_{x_j x_l} \geq \theta^2|D^2v|^2$? 
This problem is taken from Partial Differential Equations (Evans) $\S$6.4.3.

Consider the operator
$$Lu = -\sum_{i,j=1}^n a^{ij}u_{x_ix_j} +\sum_{i=1}^n b^i u_{x_i} + cu \quad\text{in }\Omega,$$
where $\Omega\subset\mathbb{R}^n$ is a bounded open set, $(a^{ij})$ is uniformly elliptic, i.e., there exists some $\theta>0$ such that
$$\sum_{i,j=1}^n a^{ij}(x)\xi_i\xi_j\geq \theta|\xi|^2$$
for any $x\in\Omega$ and $\xi\in\mathbb{R}^n$.
Given a smooth function $v$. How to prove that
$$\sum_{i,j,k,l=1}^n a^{ij} a^{kl} v_{x_i x_k} v_{x_j x_l} \geq \theta^2|D^2v|^2.$$
 A: Here is a solution, I think it's right.
Use the method as in Page 347. Fix any $x_0\in \Omega$. Since the matrix $A=(a^{ij}(x_0))$
is symmetric and positive definite, there exists an orthogonal matrix
$O=(o_{ij})$ such that
$$OAO^T = \mathrm{diag}(\lambda_1, \ldots, \lambda_n),$$
with $\lambda_i\geq \theta$ for $i=1,\ldots,n$. Write $y = x_0 + O(x-x_0)$, then
$$u_{x_ix_j} = \sum_{k,l=1}^n u_{y_ky_l} o_{ki} o_{lj}.$$
Therefore
$$\begin{align*}
  \bigstar:
  & =\sum_{i,j,k,l=1}^n a^{ij} a^{kl} v_{x_ix_k} v_{x_jx_l} \\
  & = \sum_{i,j,k,l=1}^n a^{ij} a^{kl}
      \biggl(\sum_{r,s=1}^n v_{y_ry_s} o_{ri} o_{sk}\biggr)
      \biggl(\sum_{p,q=1}^n v_{y_py_q} o_{pj} o_{ql}\biggr) \\
  & = \sum_{i,j,k,l,r,s,p,q=1}^n a^{ij} a^{kl} o_{ri} o_{sk} o_{pj} o_{ql}
      v_{y_ry_s} v_{y_py_q}.
\end{align*}$$
Note that
$$\sum_{i,j=1}^n a^{ij} o_{ri} o_{pj} = \delta_{rp} \lambda_r,$$
and
$$\sum_{k,l=1}^n a^{kl} o_{sk} o_{ql} = \delta_{sq} \lambda_s.$$
Hence
$$\begin{align*}
\bigstar
& = \sum \delta_{rp} \lambda_r \delta_{sq} \lambda_s v_{y_ry_s} v_{y_py_q} \\
& = \sum_{p,q=1}^n \lambda_p \lambda_q v_{y_py_q}^2 \\
& \geq \theta^2 |D^2 v|^2.
  \end{align*}$$
