Let $Y_i$, $i \in N$ denote iid random variables with $EY_i = 0$, $EY_i^2 = 1$. Consider the L-2 bound of the summation $\sum_{i=1}^n Y_i$.
My solution is:
Consider $F_n = \sigma(Y_1, . . . , Y_n)$ as the natural filtration, then $\sum_{i=0}^n Y_i$ is a submartingale, i.e. $E((\sum_{i=1}^{m} Y_i)^2|F_{n}) = E((\sum_{i=1}^{n} Y_i)^2|F_{n}) + (m-n)$ with $m>n$. So $E(\sum_{i=1}^{n} Y_i)^2 - E(Y_1^2)=n-1$ which implies $E(\sum_{i=1}^{n} Y_i)^2 = n$.
(Actually "i.i.d $Y_i$" directly leads to the conclusion that $E(\sum_{i=1}^{n} Y_i)^2 = n$ because $E(Y_i*Y_j)=0$, for any $i \neq j$).
Then $E((\sum_{i=1}^{n} Y_i)^4) \leq E(\sum_{i=1}^{n} Y_i)^2 * E(\sum_{i=1}^{n} Y_i)^2 = n^2$. Am I correct? Thanks!
Your computations for the moment of order two are correct.
However, for the moment of order four, the inequality $$\tag{*}\mathbb E\left[\left(\sum_{i=1}^{n} Y_i\right)^4\right] \leqslant\mathbb E\left[\left(\sum_{i=1}^{n} Y_i\right)^2\right] * \mathbb E\left[\left(\sum_{i=1}^{n} Y_i\right)^2\right] $$ does not hold in general. For example, if the sequence $(Y_i)$ is i.i.d. and $Y_i$ has a standard normal distribution, then $\sum_{i=1}^{n}Y_i$ has a centered normal distribution with variance $n$ hence the left hand side of (*) is $3n^2$ whereas the right hand side is $n^2$. However, it is possible (if of course $\mathbb E\left[Y_i^4\right]$ is finite) to find a constant $C$ such that $$ \mathbb E\left[\left(\sum_{i=1}^{n} Y_i\right)^4\right] \leqslant Cn^2. $$ You can expand the sum and group terms according to equal indexes in the summation, or proceed by induction.