TITLE: Can I replace one inequality with another?
QUESTION [7 upvotes]: Suppose I have a collection of real numbers $x_b$ where $b \in \{1, ..., n\}$, and a constant $C$ with $1/n \le C \le 1$.  Further suppose that
for all $b$, $x_b \le C \sum_{a} x_a$
Does it follow that
for all $b$, $|x_b| \le C \sum_{a} |x_a|$
Certainly if all the $x_b$ are positive then the result holds, and it suffices to check the implication when exactly one of the $x_b$ is negative.  I checked for a large random sample, but I can't get any futher than this.  I've tried comparing the $l^1$ and $l^\infty$ norms to no avail.
This problem can't be very hard to solve one way or the other.  Does any one have any ideas?

REPLY [2 votes]: Let $m = \min_i x_i$ and $M = \max_i x_i$. If $m \geq 0$ then the claim is trivial, so suppose that $m < 0$.
If $M < 0$ then we can rewrite the givens as
$$ |x_i| \geq C\sum_j |x_j|. $$
Adding all these inequalities,
$$ \sum_i |x_i| \geq nC \sum_j |x_j|, $$
and therefore $nC \leq 1$. Thus $nC=1$, and so all $x_i$ are equal, and the required inequalities trivially hold.
So the interesting case is when $m < 0 \leq M$. Intuitively, it is clear that the hardest case is when
$$x_1=m, \; x_2=\cdots=x_n=M, \; M = C\sum_i x_i = C((n-1)M + m). $$
We have to check when $$-m \leq C\sum_i |x_i| = C(\sum_i x_i - 2m) = M - 2Cm. $$
Since everything is homogeneous, we can assume that $M = 1$. This implies that
$ Cm = 1-C(n-1)$. Therefore the required inequality is
$$ -m \leq 1 -2(1-C(n-1)) = 2C(n-1) - 1. $$
Multiplying by $C$ and substituting $Cm$,
$$ C(n-1) - 1 \leq 2C^2(n-1) - C. $$
We get the quadratic inequality
$$ 0 \leq 2(n-1)C^2 - nC + 1. $$
The critical points of the quadratic are
$$ C_{+,-} = \frac{n \pm \sqrt{n^2 - 8(n-1)}}{4(n-1)}. $$
The inequality holds when $C \leq C_-$ or $C \geq C_+$. Note that roughly,
$$C_- \approx \frac{2}{n}, \; C_+ \approx 1-\frac{1}{n}. $$
When $C_- < C < C_+$, we can construct an example of the form discussed in which the required inequality fails to hold. The other case is left as an exercise to the reader.