Deducing an interpolation inequality from another interpolation inequality I'm trying to solve a problem from Evans' book "Partial differntial equations", page 291 number 10.
I've shown the following interpolation inequality:
$\int_\Omega  \vert Du \vert^2dx \leq \Big( \int_\Omega u^2 dx \Big)^{\frac{1}{2}} \cdot \Big( \int_\Omega \vert D^2u \vert^2 dx \Big)^{\frac{1}{2}}  \quad \text{for all} \quad u\in H_0^1(\Omega)\cap H^2(\Omega)$
Where $\Omega$ is a bounded domain in $\mathbb{R}^n$ with a smooth boundary, and we define:
$\vert Du \vert^2:= \sum^n_{i=1} \Big( \frac{\partial u}{\partial x_i} \Big)^2$ and $\quad \vert D^2u \vert^2:= \sum^n_{i,j=1} \Big( \frac{\partial^2 u}{\partial x_i \partial x_j} \Big)^2$
I've been asked to deduce the following interpolation inequality from this inequality (for $\infty>p\geq 2$):
$\int_\Omega  \vert Du \vert^pdx \leq \Big( \int_\Omega \vert u\vert^p dx \Big)^{\frac{1}{2}} \cdot \Big( \int_\Omega \vert D^2u \vert^p dx \Big)^{\frac{1}{2}}  \quad \text{for all} \quad u\in W _0^1(\Omega)\cap H^2(\Omega)$
And I've been given the hint of rewriting the left integral in the following way:
$\int_\Omega  \vert Du \vert^pdx  = \sum_{i=1}^n \int_\Omega \Big( \frac{\partial u}{\partial x_i} \Big)^2 \vert Du\vert^{p-2}dx $
But after repeated attempts, I have been unable to conclude how to use this to deduce the inequality. I would appreciate any help, as I don't seem to be making any progress.

Later addition:
Given a smooth compactly supported function $u$ we can say that:
$\int_{\Omega}\Big( \frac{\partial u}{\partial x_i} \Big)^2 \vert Du\vert^{p-2}dx= -\int_\Omega u \cdot \frac{\partial }{\partial x_i} \Big( \frac{\partial u}{\partial x_i} \cdot \vert Du \vert^{p-2}\Big)$
What I think the derivative of the product is:
$\frac{\partial }{\partial x_i} \Big( \frac{\partial u}{\partial x_i} \cdot \vert Du \vert^{p-2}\Big)=\frac{\partial^2 u }{\partial x_i^2} \cdot \vert Du \vert^{p-2}+ (p-2)\cdot \Big( \frac{\partial u}{\partial x_i}\Big)^2\cdot \frac{\partial^2 u}{\partial x_i^2}\cdot \vert Du\vert^{p-4}$
And since $\Big( \frac{\partial u}{\partial x_i}\Big)^2 \leq \vert Du \vert^{2}$, we conclude:
$\int_{\Omega} \Big( \frac{\partial u}{\partial x_i} \Big)^2 \cdot  \vert Du\vert^{p-2} dx \leq (p-1) \int_{\Omega} \Bigg \vert u\cdot \frac{\partial^2 u}{\partial x_i^2} \cdot \vert Du \vert^{p-2}  \Bigg\vert dx $
Also:
$\Big \vert \frac{\partial^2 u}{\partial x_i^2} \Big \vert \leq \vert D^2u \vert ^{\frac{1}{2}} $
I've thus arrived at:
$\int_{\Omega}\Big( \frac{\partial u}{\partial x_i} \Big)^2 \vert Du\vert^{p-2}dx \leq (p-1) \int_{\Omega} \Bigg \vert u\cdot \vert D^2u \vert^{\frac{1}{2}} \cdot \vert Du \vert^{p-2}  \Bigg\vert dx$
 A: There are two issues with your work above. First, it is not true that $|u_{x_ix_i}| \leq |D^2 u|^{1/2}$. Actually you have
$$|u_{x_ix_i}|\leq |D^2 u|:= \sqrt{\sum_{i,j=1}^n u_{x_ix_j}^2}.$$
Second, the product rule gives
$$\partial_{x_i} (u_{x_i}|Du|^{p-2}) = u_{x_ix_i}|Du|^{p-2} + u_{x_i}\partial_{x_i}|Du|^{p-2}.$$
Computing derivatives of norms is always a bit tricky, and it is easy to make mistakes with vector notation. It's best to write it out as
$$\partial_{x_i}|Du|^{p-2} = \partial_{x_i}(u_{x_1}^2+\cdots+u_{x_n}^2)^{(p-2)/2}.$$
Using chain rule we get 
$$\partial_{x_i}|Du|^{p-2} = (p-2)(u_{x_1}^2+\cdots+u_{x_n}^2)^{(p-4)/2}\sum_{j=1}^n u_{x_j}u_{x_jx_i} = (p-2)|Du|^{p-4}\sum_{j=1}^n u_{x_j}u_{x_jx_i}.$$
Hence, there exists a constant $C>0$ such that
$$|\partial_{x_i}|Du|^{p-2}|\leq C|D^2u||Du|^{p-2}.$$
Combining this with your previous work you get
$$\int |Du|^p\, dx \leq C\int |u||Du|^{p-2}|D^2 u|\, dx.$$
Now you have to make 2 careful applications of Holder's inequality, or one application of Generalized Holder (see Evans appendix). I'll let you stew on this, but can provide more details if needed.
