Showing that the variation in given action leads to Majorana equation This is a question on mathematics rather than the physics. It is based on QFT Q3.4, part b on Peskin and Schroeder.
My confusion stems from the fact that we are consider $\chi(x)$ to be a classical anticommuting field,that takes Grassmann numbers as their values and satisfy the following relation:
$$\alpha \beta = -\alpha \beta
\quad \text{ for any }\quad 
\alpha, \beta.$$

So I am given the action which is:
$$S = \int d^4x [\chi^\dagger i\bar{\sigma} \cdot\partial\chi + \frac{im}{2}(\chi^T\sigma^2\chi^*- \chi^\dagger \sigma^2 \chi^*)]$$
And then I am trying to show that varying this w.r.t $\chi^*$ yields Majorana equation.

My attempt:
I rewrote $S$ as:
$\large S = \int d^4x [\chi^*_a i\bar{\sigma}^\mu_{ab}\partial_\mu\chi_b + \frac{im}{2}(\chi_a\sigma^2_{ab}\chi_b- \chi^*_a \sigma^2_{ab}\chi^*_b]$
So then I found $\delta S$ and I am at the following step:
$\large \delta S = i \int d^4x [\bar{\sigma}^\mu_{ab} \delta\chi_a^*\partial_\mu\chi_b+\frac{im}{2}\sigma^2_{ab}(-\chi_a^* \delta\chi_b^*-\chi_b^* \delta \chi_a^*)] -----(i)$
Okay, so then this is how I proceeded:
$\large \delta(\chi_a^*\chi_b^*) = \chi_b^* \delta \chi_a^* + \chi_a^* \delta \chi_b^*$
But, $\large \chi_a^* \chi_b^*  = -\chi_b^* \chi_a^*$ as given to us in the question.
$\large \rightarrow \delta(-\chi_b^*\chi_a^*) = \chi_b^* \delta \chi_a^* + \chi_a^* \delta \chi_b^* $
$\large \rightarrow -\chi_a^*\delta\chi_b^*-\chi_b^*\delta\chi_a^* =\chi_b^* \delta \chi_a^* + \chi_a^* \delta \chi_b^* $
$\large \rightarrow \chi_a^*\delta\chi_b^* = -\chi_b^*\delta \chi_a^*$
But if I did the last bit correctly, then it means the terms inside the parenthesis in $\delta S$ in eqn(i) will cancel out, and it is not going to give me the Majorana equation.
Could someone help me see where I made the mistake? For this to work out, I feel like I need to get,
$\large \chi_a^*\delta\chi_b^* = \chi_b^*\delta \chi_a^*$
and I am off by a minus sign.
Edit 1: Upon thinking more, I don’t think what I am doing makes sense to me cause I feel like I am not correctly understanding what it means for $\large \chi_b^* \chi_a^* = -\chi_a^* \chi_b^*$. But i also do not know of a way out of this. So any pointer towards right direction would be very helpful.
 A: We would like to derive the majorana equation:
\begin{equation}
 i\bar{\sigma}\cdot \partial \chi - i m \sigma^2 \chi^* = 0
\end{equation}
from its action
\begin{equation}
 S = \int d^4x \left[ \chi^\dagger i \bar{\sigma} \cdot \partial \chi+ \frac{i m}{2} \left( \chi^\mathrm{T} \sigma^2 \chi - \chi^\dagger \sigma^2 \chi^* \right) \right]
\end{equation}
we use the variational approach to find the field equation such that $\delta S = 0$. Variation wrt $\chi^*$ gives
\begin{equation}
 \begin{split}
  \delta S &= S \left( \chi^* + \delta \chi^* \right) - S \left( \chi^* \right)\\
  &= \int d^4x \mathcal{L} \left( \chi^* + \delta \chi, \partial_\mu \chi^* + \partial_\mu \delta\chi^* \right) - \int d^4x \mathcal{L} \left( \chi^*, \partial_\mu \chi \right)\\
 \end{split}
\end{equation}
expand the 1st term using suffix notation:
\begin{equation}
 \begin{split}
  \int d^4x \mathcal{L} &\left( \chi^* + \delta \chi, \partial_\mu \chi^* + \partial_\mu \delta\chi^* \right)  \\ 
          &=\int d^4 x \left\{ i\left( \chi_a^* + \delta \chi_a^*\right) \bar{\sigma}_{ab}^\mu \partial_\mu \chi_b + \frac{i m}{2} \sigma_{ab}^2 \bigl[ \chi_a\chi_b - (\chi_a^* + \delta \chi_a^*)(\chi_b^* + \delta\chi_b^*) \bigr] \right\} 
 \end{split}
\end{equation}
The 2nd term in $\delta S$ can be expanded by the same token:
\begin{equation}
 \int d^4x \mathcal{L} \left( \chi^* , \partial_\mu \chi^* \right)  
 =\int d^4 x \left\{ i \chi_a^* \bar{\sigma}_{ab}^\mu \partial_\mu \chi_b + \frac{i m}{2} \sigma_{ab}^2 (\chi_{a}\chi_b - \chi_a^* \chi_b^*) \right\} 
\end{equation}
From these we simplify $\delta S$ into:
\begin{equation}
 \delta S = \int d^4x \left[ \delta\chi_a^* i\bar{\sigma}_{ab}^\mu \partial_\mu \chi_b + \frac{i m}{2} \sigma_{ab}^2(-\chi_a^*\:\delta \chi_b^* - \delta \chi_a^*\: \chi_b^*) \right]
\end{equation}
since $\sigma_{ab}^2$ is symmetric, we can rearange indices, so that
\begin{equation}
 \begin{split}
  \delta S &= \int d^4x \left[ \delta\chi_a^* i\bar{\sigma}_{ab}^\mu \partial_\mu \chi_b + \frac{i m}{2} \sigma_{ab}^2(-\delta\chi_a^*\chi_b^* - \delta \chi_a^*\: \chi_b^*) \right]\\
    &= \int d^4x \;\delta\chi_a^* \Bigl(i\bar{\sigma}_{ab}^\mu \partial_\mu \chi_b - i m \sigma_{ab}^2\chi_b^*\Bigr) 
 \end{split}
\end{equation}
we require $\delta S$ to vanish, hence we must have
\begin{equation}
 i\bar{\sigma}_{ab}^\mu \partial_\mu \chi_b - i m \sigma_{ab}^2\chi_b^* = 0
\end{equation}
