Prove this inequality: $E[(\sum_i^n{X_i})^4] \leq n \times E[X_1^4] + 3 \times n^2 \times E[X_1^2]^2$

Question

Let {$X_n$} be a sequence of i.i.d. random variables such that $E[X_1]=0$ and $E[(X_1)^4]=1.$

Prove that $E[(\sum_i^n{X_i})^4] \leq n \times E[X_1^4] + 3 \times n^2 \times E[X_1^2]^2$

Then use Borel-Cantelli to prove that

$1/n \times \sum_i^n{X_i} \rightarrow 0$ almost surely.

Answer 1

$$(\sum_{i=1}^nX_i)^4=\sum_{i=1}^nX_i^4+6\sum_{i\neq j}X^2_iX^2_j+C,$$

where $C$ contains products of $X_i$ with at least one $X_j$ occuring with a power of one: e.g. $X_1X_2^2X_3, X_1^3X_2$, etc. Use the fact that $X_i$ are independent and $E[X_i]=0$ to kill off $C$.

To see the second term notice that $X_i^2X_j^2$ is picked in $\binom{4}{2}$ ways. There are $n^2-n$

Can you finish it from here?