AM-GM Inequality Confusing Here is something that I find hard to make sense of. Suppose $X_1, X_2, ..., X_n$ are independent draws from some distribution. By AM-GM inequality, we have:
$$
\left( X_1 X_2 .. X_n \right)^\frac{1}{n} \le \frac{X_1 + X_2 +...+X_n}{n}  
$$
Now $E[X_1 X_2...X_n]=E[X_1]^n$ by independence with any $n$.
Also as $n \to \infty$, $\frac{X_1 + X_2 +...+X_n}{n} \to E[X_1]$ by law of large numbers. 
Plug these terms into the first equation, we have:
$$E[X_1] \le E[X_1]$$ when $n \to \infty$
Note that equality of the above is only possible when $X_1 = X_2 = ... = X_n$ which is highly unlikely and irrelevant.
Can someone tell me what is going on and what I am missing?
 A: By your logic on the left-hand side of this inequality (which is not actually valid, but since @zyx already explained the problems on the left-hand side, I'm going to assume that it is valid to explain why your reasoning on the right side is invalid), the following is true for any $n$ with the situation you gave:
$$E[X_1] \leq \frac{X_1+X_2+X_3+...+X_n}{n}$$
You then say that $\frac{X_1+X_2+X_3+...+X_n}{n} \to E[X_1]$ as $n \to \infty$. That's true, but using that, you can not substitute $E[X_1]$ for the average because that only becomes true as $n \to \infty$. It's not necessarily true that $\frac{X_1+X_2+X_3+...+X_n}{n}=E[X_1]$ for any $n$, but rather that, as $n$ gets larger, the first value becomes closer to the second value.
What you can say is the original inequality at the top of this answer and that as $n \to \infty$, the right-hand side becomes closer to $E[X_1]$. The right-hand side approaches the left-hand side as $n \to \infty$. That is not the same as saying that $E[X_1] \leq E[X_1]$ as $n \to \infty$ with equality only if all $X_i$ are equal, which simply does not make any sense.
A: The expectation of $\left(\frac{X_1 + X_2 +...+X_n}{n} \right)^n$ is not $E[X]^n$.  For $n=2$ as an example, $$E[(X+Y)^2/4] = E[(X^2 + Y^2 + 2XY)/4] = (E[X^2] + E[X]^2)/2 = E[X]^2 + \frac{1}{2}{\rm Var}(X) \geq E[X]^2$$ if $X$ and and $Y$ are independent samples from the same distribution.

In the Edited version, the same problem moves to the other side of the claimed equation; $E[\left( X_1 X_2 .. X_n \right)^\frac{1}{n}] = E[X^{\frac{1}{n}}]^n$ is not $E[X]$.  
