Inequality of the laplacian involving the Ricci curvature

I am reading Eschenburg and Heintze's proof of the Cheeger-Gromoll splitting theorem. Lemma 1 states:

Let $$f\in C^\infty(M)$$ with $$||grad(f)||=1$$. If c is an integral curve of the gradient, then it is a minimising geodesic and \begin{align} -Ricc(c', c') & =(\Delta f\circ c)'+||Hess_f\circ c||^2 \\ & \le (\Delta f\circ c)' +\frac{1}{n-1}(\Delta f\circ c)^2 \end{align}

They begin the proof by choosing some $$t_0\in \mathbb{R}$$ and they take a neighbourhood $$U$$ of $$c(t_0)$$, where they choose and orthonormal frame $$\{ E_1,E_2,...,E_n\}$$ such that $$E_n=gradf$$ and $$E_i$$ are parallel along $$gradf$$. Then the calculation give \begin{align} Ricc(E_n,E_n) &=\sum_{i=1}^{n} \left< R(E_i,E_n)E_n,E_i\right> \\ &=\sum_i\left( -\left< \nabla_{E_n}\nabla _{E_i}E_n,E_i \right> -\left< \nabla_{\nabla_{E_i}E_n}E_n,E_i\right> \right) \\ &=-E_n\left(\sum_i \left< \nabla_{E_i}E_n,E_i\right>\right) -\sum_i\left< \nabla_{E_i}E_n,\nabla_{E_i}E_n\right> \\ &=-E_n(\Delta f)-||Hess_f||^2 \end{align}

Tis way we prooved th equality of the Lemma. I understand intuitively the existence of such a frame $$\{ E_1,E_2,...,E_n\}$$ but I would like to see a rigorous proof. Mainly I don't understand how one gets from this to the inequality. All the authors say is that from the Schwarz inequality we have \begin{align} Ricc(E_n,E_n) & \leq -E_n(\Delta f)-\sum_{i=1}^{n-1}\left\\ & \leq -E_n(\Delta f) -\frac{1}{n-1}(\Delta f)^2 \end{align} But I can't see how it is done

For the second question, note that $$\operatorname{Hess}f$$ is a symmetric matrix and thus can be diagonalized by some orthonormal basis $$\{E_1, \cdots, E_n\}$$. We can also choose $$E_n = c' = \nabla f$$ since $$c'$$ is an eigenvector with eigenvalue $$0$$: $$\operatorname{Hess} f(c') = c' c' f - \nabla_{\nabla_{c'}c'} f = 0$$
$$\operatorname{Hess} f(E_i, E_j) = \delta_{ij} \lambda_i.$$
with $$\lambda_n = 0$$. In terms of $$\lambda_i$$'s we have $$\Delta f = \sum_{i=1}^{n-1} \lambda_i, \ \ \ \| \operatorname{Hess} (f)\|^2 = \sum_{i=1}^{n-1} \lambda_i^2.$$
\begin{align} \Delta f &= \sum_{i=1}^{n-1} \lambda_i\\ &= (\lambda_1, \cdots, \lambda_{n-1}) \cdot (1, \cdots, 1) \\ &\le |(\lambda_1, \cdots, \lambda_{n-1})| | (1, \cdots, 1)| \\ &= \sqrt{\lambda_1^2 + \cdots + \lambda_{n-1}^2} \sqrt{n-1} \\ \Rightarrow (\Delta f)^2 &\le (n-1) \|\operatorname{Hess} f\|^2 \end{align}