Use of Holder inequality in gradient estimate for harmonic function. While reading the book "Elliptic Partial Differential Equations" by Han and Lin, I failed to understand the proof of the interior gradient estimate for harmonic functions. The theorem says that if $u$ is harmonic in $B_1 (\subset \mathbb{R}^n)$, then $ \sup_{B_{1/2}}{|Du|} \le c\sup_{\partial B_1}{|u|}$ for some positive constant $c=c(n).$ The proof illustrated in the book begins with choosing a cut-off function $\phi = \eta^2$ for some $\eta \in C_{0}^{1}(B_1)$ with $\eta = 1$ in $B_{1/2}.$ Directly calculating, one can show $$\Delta(\eta^2|Du|^2) = 2\eta\Delta\eta|Du|^2 + 2|D\eta|^2|Du|^2 + 8\eta\sum_{i, j =1}^{n}{D_{i}\eta D_{j}u D_{ij}u} + 2\eta^2\sum_{i,j=1}^{n}(D_{ij}u)^2.$$ What I do not know is the following inequality: $\Delta(\eta^2|Du|^2) \ge (2\eta\Delta\eta-6|D\eta|^2)|Du|^2.$ The book explains that "Holder inequality" is used here, but the only thing that I know regarding to Holder inequality is what appears in the usual real analysis book in the chapter of $L^p$ spaces (inequality for integral). I guess by the form of the equation, we need to say something about the third term: $8\eta\sum_{i, j =1}^{n}{D_{i}\eta D_{j}u D_{ij}u}.$
To be specific, obtaining the inequality $8\eta\sum_{i, j =1}^{n}{D_{i}\eta D_{j}u D_{ij}u}  \ge -8|D\eta|^2|Du|^2$ will end the proof.
Can anyone help me to see how Holder inequality could be used to prove the claimed inequality? Thanks in advance.
 A: From googling, I managed to end up at these MIT notes that expand slightly on the proof (but don't use Holder's). The trick is as follows. Note that it is enough to show
$$ 8 \eta \sum_{ij} D_i \eta D_j u D_{ij} u + 2\eta^2 \sum_{ij} (D_{ij}u)^2 \ge - 8 |D\eta |^2 |Du|^2$$
Note that in each term, $\eta$ and $D_{ij}u$ appear in equal powers. So define
$$ X_{ij} = \eta D_{ij} u ,\quad Y_{ij} = D_i \eta D_j u$$
Then the above inequality is equivalent to
$$ \sum_{ij} 4X_{ij}Y_{ij} + 4Y_{ij}^2 + X_{ij}^2  \ge 0$$
but of course this LHS is nothing but
$$ \sum_{ij} 2X_{ij}(2Y_{ij}) + (2Y_{ij})^2 + X_{ij}^2 = \sum_{ij} (2Y_{ij}+X_{ij})^2$$
which is clearly non-negative, so the original inequality that we wanted to prove is also true.
For vectors, the inequality $|a+b|^2\ge 0$ can be used to prove Cauchy-Schwarz ($\sum_i a_ib_i \le |a||b|$) so I can imagine there is a version of this proof that uses Cauchy-Schwarz...but I don't see it.
In addition, this book (A Basic Course in Partial Differential Equations) by Han expands on the proof in a different way, using the Cauchy inequality (i.e. Young's inequality for products, a.k.a. the "rob-Peter-to-pay-Paul" inequality; perhaps this is a typo in Han and Lin!).
Screenshot for proof from Google Books, and a transcription:

This is the global gradient estimate. To get interior estimates, we need to introduce a cutoff function. For any nonnegative function $\varphi \in C_{0}^{2}\left(B_{1}\right),$ we have
$$
\Delta\left(\varphi|\nabla u|^{2}\right)=(\Delta \varphi)|\nabla u|^{2}+4 \sum_{i, j=1}^{n} \varphi_{x_{i}} u_{x_{j}} u_{x_{i} x_{j}}+2 \varphi \sum_{i, j=1}^{n} u_{x_{i} x_{j}}^{2}
$$
By the Cauchy inequality, we get
$$
4\left|\varphi_{x_{i}} u_{x_{j}} u_{x_{i} x_{j}}\right| \leq 2 \varphi u_{x_{i} x_{j}}^{2}+\frac{2}{\varphi} \varphi_{x_{i}}^{2} u_{x_{j}^{*}}^{2}
$$
Then
$$
\Delta\left(\varphi|\nabla u|^{2}\right) \geq\left(\Delta \varphi-\frac{2|\nabla \varphi|^{2}}{\varphi}\right)|\nabla u|^{2}
$$
We note that the ratio $|\nabla \varphi|^{2} / \varphi$ makes sense only when $\varphi \neq 0 .$ To interpret this ratio in $B_{1},$ we take $\varphi=\eta^{2}$ for some $\eta \in C_{0}^{2}\left(B_{1}\right) .$ Then
$$
\Delta\left(\eta^{2}|\nabla u|^{2}\right) \geq\left(2 \eta \Delta \eta-6|\nabla \eta|^{2}\right)|\nabla u|^{2} \geq-C|\nabla u|^{2}
$$
where $C$ is a positive constant depending only on $\eta$ and $n .$ Note that
$$
\Delta\left(u^{2}\right)=2|\nabla u|^{2}+2 u \Delta u=2|\nabla u|^{2}
$$

