Cheeger-Gromoll splitting proof Background information: $E_i$ is the parallel orthonormal frame along $c$ and $E_n=\nabla f\circ c$.
Lemma: Let $M$ be a Riemannian manifold and $f\in C^\infty(M)$ with $||\text{grad} f||=1$. If $c$ is an integral curve of $\text{grad}f$, then $c$ is a geodesic realizing the distance between any two of its points and
$$\begin{align}
-\text{Ric}(\dot{c},\dot{c})&=(\Delta f\circ c)'+||\text{Hess}f\circ c||^2\\
&\geq (\Delta f\circ c)'+\frac{1}{n-1}(\Delta f\circ c)^2\\
\end{align}
$$
where $\Delta f=\text{tr} \text{ Hess} f$.
Proof

This is a proof from Eschenburg and Heintze's paper, An Elementary Proof of the Cheeger-Gromoll Splitting theorem. Below are my questions and concerns. I would really appreciate it if someone could help me understand the steps. Thanks.
Question 1: Using Petersons' text it seems that $$||\text{Hess}f||^2=\sum_{i,j=1}^n\langle \nabla_{E_i}E_n,E_j\rangle^2$$
so I don't see how the author gets his computation from. 
Question 2 I don't see why $-||\text{Hess}f||^2\leq- \sum_{i=1}^n\langle\text{Hess}f(E_i),E_i\rangle^2$
Question 3 I don't see how they get the inequality $-\sum_{i=1}^n\langle\text{Hess}f(E_i),E_i\rangle^2\leq -\frac{1}{n-1}(\Delta f)^2$.
 A: Question 1 Given a function $f : M \to \mathbb R$ and $g$ a metric, the Hessian of $f$, $H = \text{Hess}_f$ is a $(1,1)$-tensor $TM \to TM$ defined as 
$$ \text{Hess}_f (X)  =\nabla_X \nabla f.$$
For any $(1,1)$ tensor, one can define the norm $\|H\|$ to be 
$$\| H\|^2 = \sum_{i,j=1}^n \langle H(E_i), E_j\rangle^2,$$
where $\{E_1, E_2, \cdots, E_n\}$ is any orthonormal basis. Thus in our situation
$$\|\text{Hess}_f\|^2 = \sum_{i,j=1}^n \langle \text{Hess}_f(E_i), E_j\rangle^2 = \sum_{i,j=1}^n \langle \nabla_{E_i} \nabla f , E_j\rangle^2 = \sum_{i,j=1}^n \langle \nabla_{E_i} E_n , E_j\rangle^2$$
as in our case $\nabla f = E_n$ along $c$. 
Question 2 follows from the defintion: 
$$\begin{split} \|\text{Hess}_f\|^2 &= \sum_{i,j=1}^n \langle \text{Hess}_f(E_i) , E_j\rangle^2 \\
&\ge \sum_{i=1}^n \langle \text{Hess}_f (E_i), E_i\rangle^2.
\end{split}$$
Question 3 Using $\Delta f = \text{tr}\text{Hess}_f$, 
$$\begin{split}
(\Delta f)^2 &= (\text{tr} \text{Hess}_f)^2 \\
&= \left(\sum_{i=1}^n \langle\text{Hess}_f(E_i), E_i\rangle\right)^2 \\
&=  \left(\sum_{i=1}^{n-1} \langle\text{Hess}_f(E_i), E_i\rangle\right)^2
\end{split}$$
since the last term is zero:
$$\langle \text{Hess}_f (E_n), E_n \rangle= \langle \nabla_{E_n} E_n,E_n\rangle =\frac 12 E_n \langle E_n, E_n\rangle = 0.$$
Now by Cauchy Schwarz inequality, 
$$\begin{split} (\Delta f)^2 &= \left(\sum_{i=1}^{n-1} \langle\text{Hess}_f(E_i), E_i\rangle\right)^2 \\
&\le \left(\sum_{i=1}^{n-1} 1^2 \right)\left(\sum_{i=1}^{n-1} \langle \text{Hess}_f (E_i), E_i\rangle^2 \right)\\
&= (n-1) \left(\sum_{i=1}^{n-1} \langle \text{Hess}_f (E_i), E_i\rangle^2 \right).
\end{split}$$
