**Question:** Suppose we have $N$ independent random variables $X_1$, $\ldots$, $X_N$ with finite means $\mu_1 \leq \ldots \leq \mu_N$ and variances $\sigma_1^2$, $\ldots$, $\sigma_N^2$.

I am looking for distribution-free bounds on the probability that any $X_i \neq X_N$ is larger than all other $X_j$, $j \neq i$.

In other words, if for simplicity we assume the distributions of $X_i$ are continuous (such that $P(X_i = X_j) = 0$), I am looking for upper bounds for: $$ P( X_i = \max_j X_j ) \enspace. $$ More specifically, I want to lower bound: $$ \sum_{i=1}^N \mu_i P( X_i = \max_j X_j ) \enspace. $$ Below is a possible answer, but the bound seems quite loose and I would like to know if sharper bounds can be formulated (ideally without additional assumptions).

Please note that the variables are **not** assumed to be i.i.d..

**My attempt:** Recall that by assumption, $\mu_j \geq \mu_i$ whenever $j > i$. If $N=2$, we can use Chebyshev's inequality to get
$$
P(X_1 = \max_j X_j) = P(X_1 > X_2) \leq \frac{\sigma_1^2 + \sigma_2^2}{\sigma_1^2 + \sigma_2^2 + (\mu_1 - \mu_2)^2} \enspace.
$$
We can use this bound for general $N \geq 2$ to arrive at
$$
P(X_i = \max_j X_j) \leq \min_{j > i} \frac{\sigma_i^2 + \sigma_j^2}{\sigma_i^2 + \sigma_j^2 + (\mu_j - \mu_i)^2} \leq \frac{\sigma_i^2 + \sigma_N^2}{\sigma_i^2 + \sigma_N^2 + (\mu_N - \mu_i)^2} \enspace.
$$
This implies, for all $i$
$$
( \mu_N - \mu_i ) P( X_i = \max_j X_j ) \leq (\mu_N - \mu_i) \frac{\sigma_i^2 + \sigma_N^2}{\sigma_i^2 + \sigma_N^2 + (\mu_N - \mu_i)^2} \leq \frac{1}{2} \sqrt{ \sigma_i^2 + \sigma_N^2 } \enspace.
$$
This, in turn, implies
$$\tag{1}
\sum_{i=1}^N \mu_i P( X_i = \max_j X_j ) \geq \mu_N - \frac{N-1}{2} \sqrt{ \sum_{i=1}^{N-1} (\sigma_i^2 + \sigma_N^2) } \enspace.
$$
I am trying to find out whether this bound can be improved to something that does not depend linearly on $N$. For instance, does the following hold:
$$\tag{2}
\sum_{i=1}^N \mu_i P( X_i = \max_j X_j ) \geq \mu_N - \sqrt{ \sum_{i=1}^N \sigma_i^2 } \enspace?
$$
And if not, what could be a counterexample?