Discrete Hölder inequality I am reading the proof of Proposition 1.31 in "Elliptic Partial Differential Equations" by Han & Lin.
It must be an easy application of the discrete Hölder inequality, but I don't get it. 
Let $u$ be an harmonic function defined in the unitary ball $B_1 \subset \mathbb{R}^n$ and let $\eta \in C^1_0(B_1)$ be a cutoff function such that $\eta \equiv 1$ in $B_{\frac{1}{2}}$.
Then by a direct calculation we have:
\begin{align}
\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.
\end{align}
Then the book says that from the Hölder inequality we have:
\begin{align}
\Delta ( \eta^2 |Du|^2) &\ge (2 \eta \Delta \eta -6 |D\eta|^2)|Du|^2.
\end{align}
Probably it's trivial, but I can't understand why. 
 A: I found the solution to my problem. Indeed it was just an application of the Hölder inequality and of the inequality $a^2 + b^2 \ge 2ab$ as suggested in the comment. 
I showed that:
$$
8\eta \sum_{ij=1}^n D_i \eta D_j u D_{ij}u \ge - \Big( 8 |D\eta|^2|Du|^2 + 2 \eta^2 \sum_{ij=1}^n( D_{ij} u)^2 \Big)
$$
and this is sufficient to get the statement of the book. 
Here is the computation. 
I actually proved the inequality above multiplied by $-1$.
\begin{align}
-8 \eta \sum_{ij=1}^n D_i \eta D_j u D_{ij} u &\le 8 \sum_{ij=1}^n | \eta D_i \eta D_j u D_{ij}u| \\
&\le 8\sum_{i = 1}^n \big(|D_i \eta| \sum_{j=1}^n | \eta D_j u D_{ij} u| \big)\\
&\overset{Hölder}{\le} 8|D\eta|\Big( \sum_{i=1}^n\Big(\sum_{j=1}^n | \eta D_j u D_{ij} u|\Big)^2\Big)^{\frac{1}{2}} \\
&\overset{Hölder}{\le} 8|D\eta||Du| |\eta|\Big(\sum_{ij=1}^n(D_{ij}u)^2\Big)^{\frac{1}{2}}
\end{align}
Now we use the inequality  $ab \le \frac{a^2}{2} + \frac{b^2}{2}$ with $a = 4|D\eta||Du|$ and $b = 2 |\eta| \sum_{ij=1}^n(D_{ij}u)^2$ and we get the conclusion. 
