Assume $X_1,X_2,\ldots,X_N$ are identically distributed, non-negative, not independent. If $E\left[\sum\limits_{k=1}^N \frac{X_k}{N}\right]=\mu$, is there a way that I can conclude $$\lim\limits_{N \to \infty}P\left(\sum\limits_{k=1}^N \frac{X_k}{N}>\frac{\mu}{2}\right)=1$$
If $X_i$s are independent, we can use the weak law of large numbers. But, how about the non-independent version?
I know one counter example is to set $X_1=X_2=\cdots=X_N$. Then, $$\lim\limits_{N \to \infty}P\left(\sum\limits_{k=1}^N \frac{X_k}{N}>\frac{\mu}{2}\right)=\lim\limits_{N \to \infty}P\left( X_k>\frac{\mu}{2}\right)$$ which is not necessaruly true.
Is there any condition to add which makes it possible to prove what I need? i.e, $\lim\limits_{N \to \infty}P\left(\sum\limits_{k=1}^N \frac{X_k}{N}>\frac{\mu}{2}\right)=1$?