Suppose that $\epsilon_1,\ldots,\epsilon_n$ is a sequence of independent random variables, each taking the values $1,-1$ both with probability $1/2$. If $\alpha_1,\ldots,\alpha_n$ is a sequence of real numbers, define a random variable $Z$ and $\sigma\ge 0$ by,
$$Z = \sum_{k=1}^n\epsilon_k\alpha_k,\ \sigma^2 = \sum_{k=1}^n\alpha_k^2$$ so that $Z$ has mean zero and variance $\sigma^2$.
How can we show that
$$E[Z^4] \le 3\sigma^4$$
Let $b_k=a_k\epsilon_k$. When you expand out $$ Z^4=\left(\sum b_k\right)^4, $$ you get terms like $$ b_i^4,\quad b_i^2b_j^2,\quad b_i^3b_j,\quad b_i^2b_jb_k,\quad b_ib_jb_kb_\ell, $$ where $i,j,k,\ell$ are distinct indices. The expectations of each of these (using the fact that the variables are independent with zero mean) are equal to $$ a_i^4,\quad a_i^2a_j^2,\quad 0,\quad 0,\quad0. $$ Therefore, $$ E[Z^4] = \sum_ia_i^4+3\sum_{i\neq j}a_i^2a_J^2\le 3\sum_{i,j}a_i^2a_j^2= 3(\sigma^2)^2 $$