Following Aubin's book "Some nonlinear problems in Riemannian geometry", we use the notation
$$
|\nabla^r \psi|^2 = \nabla\_{\alpha\_1}\cdots \nabla\_{\alpha\_r}\psi \nabla^{\alpha\_1}\cdots \nabla^{\alpha\_r}\psi
$$
where $\nabla\_\alpha$ is the covariant derivative and $\nabla^\alpha :=g^{\alpha\beta} \nabla\_\beta$.


The statement is the first part of the proof of **proposition 2.11** on page 36. For simplicity, let me quote the statement for $r=1$ of the proposition:



> 
> Let $r=1$ and let $\psi\in C^{r+1}(M)$, then 
>  $$
> |\nabla |\nabla^r \psi|| \le |\nabla^{r+1} \psi|.
> $$
>  To establish this inequality, it is sufficient to develop 
>  $$
> (\nabla\_\nu \nabla\_\alpha \psi \nabla\_\beta\psi - \nabla\_\nu\nabla\_\beta\psi\nabla\_\alpha\psi)
> \times g^{\nu\mu} g^{\alpha \lambda} g^{\beta\gamma} 
> (\nabla\_\mu \nabla\_\lambda \psi \nabla\_\gamma\psi - \nabla\_\mu\nabla\_\gamma\psi\nabla\_\lambda\psi) 
> \ge 0.
> $$
>  We find $4|\nabla^{2} \psi|^2|\nabla \psi|^2 - |\nabla |\nabla \psi|^2|^2 \le 0$.
> 
> 
> 


I must admit that I don't understand his line of deduction at all. Why is establishing the long inequality help? How do we deduce/use the last inequality?


Any help is very appreciated.



Proving the inequality $|\nabla |\nabla^r \psi|| \le |\nabla^{r+1} \psi|$
The line Aubin says to "develop" is the inner product of the $(0, 3)$ tensor $T\_{\nu\alpha\beta} = \nabla\_\nu \nabla\_\alpha \psi \nabla\_\beta \psi - \nabla\_\nu \nabla\_\beta \psi \nabla\_\alpha \psi$ with itself, i.e. $|T|^2\_{g}$. That's why it's positive.


By "develop" Aubin means "distribute out the multiplication", which lands us at
\begin{align}
|T|^2\_{g} 
&= 
g^{\nu \mu}g^{\alpha \lambda}g^{\beta \gamma}
\nabla\_\nu \nabla\_\alpha \psi \nabla\_\beta \psi \nabla\_{\mu}\nabla\_\lambda\psi\nabla\_{\gamma}\psi \\
&-
g^{\nu \mu}g^{\alpha \lambda}g^{\beta \gamma}
\nabla\_\nu \nabla\_\alpha \psi \nabla\_\beta \psi 
\nabla\_\mu\nabla\_\gamma \psi \nabla\_\lambda \psi \\
&-
g^{\nu \mu}g^{\alpha \lambda}g^{\beta \gamma}
\nabla\_\nu \nabla\_{\beta} \psi \nabla\_\alpha {\psi}
\nabla\_{\mu}\nabla\_\lambda\psi\nabla\_{\gamma}\psi \\
&+
g^{\nu \mu}g^{\alpha \lambda}g^{\beta \gamma}
\nabla\_\nu \nabla\_{\beta} \psi \nabla\_\alpha {\psi}
\nabla\_\mu\nabla\_\gamma \psi \nabla\_\lambda \psi.
\end{align}
We can recognize the first and last lines as $|\nabla \nabla \psi|^2|\nabla \psi|^2$.


For the other lines, it's easier to do the calculation backwards: note
\begin{align}
|\nabla |\nabla \psi|^2|^2
&= g^{ab}
\nabla\_a(g^{cd}\nabla\_c\psi\nabla\_d\psi) 
\nabla\_b(g^{ef}\nabla\_e\psi\nabla\_f\psi)\\
\end{align}
since I can switch $c,d$ in the inverse-metric we get the same term when the $\nabla\_a$ hits either $\nabla\_c$ or $\nabla\_d$. Similarly with $\nabla\_b$. So
\begin{align}
|\nabla |\nabla \psi|^2|^2
&= 4g^{ab}g^{cd}g^{ef}
(\nabla\_a\nabla\_c\psi\nabla\_d\psi) 
(\nabla\_b\nabla\_e\psi\nabla\_f\psi)\\
\end{align}
Now, you can match the terms above with the two terms which come with a minus sign in our big calculation of $|T|^2\_g$. We find,
$$|T|^2\_g = 2|\nabla \nabla \psi|^2 |\nabla \psi|^2 - (1/2)|\nabla |\nabla \psi|^2|^2.$$
This proves the claim at the end.


To use the claim, let $f = |\nabla \psi|$ and note
$$|T|^2\_g = 2|\nabla \nabla \psi|^2 f^2 - (1/2)|\nabla f^2|^2$$
Now do the derivative, divided by $f^2$, and find what you want.


If you're reading this inequality for the first time, it may be useful to note the exact constant of $1$ in the inequality is sometimes important in geometric analysis for elliptic pde. It comes up when calculating the laplacian of norms of $u$ or norms of derivatives of $u$, assuming you know something about the laplacian of $u$. Say $u$ is a tensor, if we calculate
$\Delta |u|^2$ we get a positive term $2|\nabla u|^2$. If you then use that to calculate $\Delta |u|$ you pick up a negative term (since $|u| = \sqrt{|u|^2}$ and $x \mapsto \sqrt{x}$ is concave) which is always beat by the positive term $2|\nabla u|^2$, using the inequality. 


