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.