One can use Birnbaum and Marshall inequality:
Theorem(Theorem 2.1. in 1). If $\left(S_k,k\geqslant 1\right)$ is a non-negative sub-martingale and $(c_k,k\geqslant 1)$ a non-decreasing sequence of positive numbers, then for each $p\geqslant 1$:
$$\mathbb P\left\{\max_{1\leqslant k\leqslant n}\frac{S_k}{c_k}\geqslant 1\right\}\leqslant \frac{\mathbb E\left[S_n^p\right]}{c_n^p}+\sum_{i=1}^{n-1}\left(\frac 1{c_i^p}-\frac 1{c_{i+1}^p}\right)\mathbb E\left[S_i^p\right].$$
Reference:
1 [Some Multivariate Chebyshev Inequalities with Extensions to Continuous Parameter Processes]1
Z. W. Birnbaum and Albert W. Marshall, Ann. Math. Statist. Volume 32, Number 3 (1961), 687-703.
To complement my other answer, I will show
Proposition 1 Let $\xi_k$ be a finite number of iid Bernoulli random variables of expectation $p > 1/2$, and let $a_k > 0$ be real numbers. Then ${\bf P}( \sum_k a_k \xi_k \geq p \sum_k a_k) \gg 1$.
By replacing $\xi_k$ with $1-\xi_k$ and $p$ with $1-p$, this is equivalent to
Proposition 2 Let $\xi_k$ be a finite number of iid Bernoulli random variables of expectation $p < 1/2$, and let $a_k > 0$ be real numbers. Then ${\bf P}( \sum_k a_k \xi_k \leq p \sum_k a_k) \gg 1$.
Let's prove Proposition 2. I found a number of arguments to treat various special cases, which when combined together was able to cover the general case as follows.
We will need a large absolute constant $C_0$.
Case 1: (high multiplicity) For every integer $n$, the number of $a_k$ in the dyadic interval $(2^{n-1}, 2^n]$ is either zero, or at least $C_0/p$.
Roughly speaking this is the regime where the central limit theorem applies.
In this case we can proceed by the Berry-Esseen inequality, which lets one estimate
$$ {\bf P}( \sum_k a_k \xi_k \geq p \sum_k a_k) = \frac{1}{2} + O( \frac{p \sum_k a_k^3}{(p \sum_k a_k^2)^{3/2}} ).$$
If, for each $n$, we let $c_n$ be the number of $k$ for which $a_k \in (2^{n-1},2^n]$, we can write the right-hand side as
$$ \frac{1}{2} + O( p^{-1/2} \frac{\sum_n 2^{3n} c_n}{(\sum_n 2^{2n} c_n)^{3/2}} ).$$
By hypothesis, each $c_n$ is either zero or at least $C_0/p$, so we can bound
$$ 2^{3n} c_n \leq (C_0/p)^{-1/2} 2^{2n} c_n (\sum_{n'} 2^{2n'} c_{n'})^{1/2} $$
so the previous expression becomes $\frac{1}{2} + O( C_0^{-1/2} )$, and the claim follows for $C_0$ large enough.
Case 2: (low multiplicity) For every integer $n$, the number of $a_k$ in the dyadic interval $(2^{n-1}, 2^n]$ is at most $C_0/p$.
Here the Bernoulli variables are mostly zero and one can proceed using the first and second moment methods, after first applying a dyadic decomposition.
Let $n_0$ be the integer such that $p \sum_k a_k \in (2^{n_0-1}, 2^{n_0}]$. It will suffice to show that
$$ {\bf P}( \sum_k a_k \xi_k \leq 2^{n_0-1} ) \gg_{C_0} 1.$$
The number of $k$ with $a_k > 2^{n_0-C_0}$ is at most $2^{C_0} / p$, and hence
$$ {\bf P}( \sum_{k: a_k > 2^{n_0-C_0}} a_k \xi_k = 0 ) \geq (1-p)^{2^{C_0}/p} \gg_{C_0} 1.\qquad(1)$$
Next, for any $m \geq C_0$, we consider the random variable
$$ \sum_{k: 2^{n_0-m-1} < a_k \leq 2^{n_0-m}} a_k \xi_k.$$
There are at most $C_0/p$ elements in this sum, so the second moment of this variable can be computed as
$$ {\bf E} ( \sum_{k: 2^{n_0-m-1} < a_k \leq 2^{n_0-m}} a_k \xi_k )^2 \ll C_0^2 2^{2n_0-2m}.$$
By Markov's inequality we thus have
$$ {\bf P} ( \sum_{k: 2^{n_0-m-1} < a_k \leq 2^{n_0-m}} a_k \xi_k \geq 2^{n_0-m/2} ) \ll C_0^2 2^{-m}.\qquad(2)$$
Applying the union bound to (2) for all $m \geq C_0$ and combining with (1) (using independence), we obtain the claim (if $C_0$ is large enough).
Case 3: (general case)
In this case we can partition the $k$ into two classes ${\mathcal K}_1$, ${\mathcal K}_2$, one of which is in the low multiplicity case and one of which is in the high multiplicity case. By the preceding cases we have
$${\bf P}( \sum_{k \in {\mathcal K}_1} a_k \xi_k \leq p \sum_{k \in {\mathcal K}_1} a_k) \gg 1$$
and
$${\bf P}( \sum_{k \in {\mathcal K}_2} a_k \xi_k \leq p \sum_{k \in {\mathcal K}_2} a_k) \gg_{C_0} 1$$
so by independence we conclude
$${\bf P}( \sum_k a_k \xi_k \leq p \sum_k a_k) \gg_{C_0} 1$$
as required.
Best Answer
It is always true. Split $x_i=y_i+z_i$ where $y_i$ are bounded and $Ez_i\le \frac \mu{10n}$. You have no problems with $y_i$ because if they were alone,$ES_j$ would be concentrated in a very strong sense around $\nu T$ for large $T$ where $\nu=Ey_i\le\mu$ (see Didier's argument for details or recall the Bernstein inequality for exponential moments of sums of bounded variables). But you also have no problems with $z_i$ because even if you add them all up instead of taking the maximum at some point, you end up with mere $0.1 T\mu$.