The inequality is false in general. A counterexample:
Let $n = 7$. Let $x_1 = -1$. Let $x_2 = \cdots = x_7 = \frac{\sqrt{2}+1}{6}$. And let $C = \frac{\sqrt{2}+1}{6\sqrt{2}}$. Then $ \sum x_a = -1 + \sqrt{2} + 1 = \sqrt{2} $ so $C\sum x_a \geq x_b$ for any $b$.
But $ \sum |x_a| = \sqrt{2} + 2 $ so $ C\sum |x_a| = \frac{3+2\sqrt{2}}{6} < 1 = |x_1|$
Interestingly, for $n \leq 6$ the inequality is true. A sketch of proof goes something like this:
The inequality is trivially true if all of $x_a$ are non-negative, or if $C = 1/n$ (since in that case we must have all the $x_a$ are identical). So we assume that $C>1/n$, and that there are $k>0$ elements which are negative. We can assume that those are the first $k$. Then the given inequality implies $ x_b \leq C \sum_{k+1}^n x_a - C \sum_1^k |x_a| $ we sum $b$ from $k+1$ to $n$, and we have that $ C(n-k)\sum_1^k |x_a| \leq [C(n-k)-1] \sum_{k+1}^n x_a $ notice that this puts a restriction on $k$ such that $C(n-k)-1 > 0$. Denote by $\delta = C(n-k)-1$. We can re-write the above as $ (2\delta + 1)\sum_1^k |x_a| \leq \delta \sum_1^n |x_a| $
So if $C(2\delta + 1)/\delta \geq 1$, the above expression will imply the desired inequality. But $C = (1+\delta)/(n-k)$, so we have $\frac{C(2\delta + 1)}{\delta} = \frac{\sqrt{2}}{n-k}(\frac{1}{\sqrt{2}\delta} + \sqrt{2}\delta) + \frac{3}{n-k} $ Using that $x + x^{-1} \geq 2$, we have that $ \geq \frac{2\sqrt{2} + 3}{n-k} $
So if $n \leq 6$, the above is guaranteed to be at least 1 (since $k \geq 1$ and 2\sqrt{2} > 2).
Remark Interestingly, the crucial number is $n-k$. If a priori you know that no more than 5 of the $x_a$ are positive, then the desired inequality will hold.
A few additional comments:
- Yes, I did the proof first before writing down the example. Hence the near-optimum constants.
- We see that the constraint on $n-k$ can be relaxed if we know a priori that $\delta = C(n-k) - 1 \leq n-k-1$ is large. Of course, in the case that $C = 1$, the conclusion is a triviality for any $n$. A computation shows the following crude asymptotic lower-bound: the inequality is true whenever $n < \frac{3 - \sqrt{2}}{1-C}$ (the inequality can be true even for some $n$ bigger than that; but as $C\to 1$ from below, this should give the right grow rate).