Suppose I have a sequence of $n$ i.i.d. random variables $X_1,X_2,\ldots,X_n$, such that they are positive $X_i>0$ and have variance $\sigma(n)=\omega(1)$ that is increasing in $n$ while the mean $\mu$ is constant.  I am trying to lower-bound the probability that the maximum of this sequence $X_\max$ exceeds a threshold $T$ (which may either be fixed or be a function of $n$ that grows slower that $\sigma(n)$).

By elementary probability we have:

$$P(X_\max> T)=1-(P(X_1\leq T))^n=1-(1-P(X_1>T))^n$$

Thus, I want to lower-bound the probability that a positive random variable $X_1$ with increasing variance exceeds a threshold.  Unfortunately, the only variance-based bound that I know is Chebyshev's and it's an upper bound.  I also know that most extreme value theory results rely on the structure of the particular distribution function...  

However, intuitively, it seems that, since $X_1$ is positive, then if the variance of $X_1$ is growing faster than the threshold, $X_1$ should exceed $T$ with high probability...  But I am having hard time proving this...  can anyone help?