Proving the inequality $|\nabla |\nabla^r \psi|| \le |\nabla^{r+1} \psi|$ 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.
 A: 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.  
