Subfield: Algebra


Question: For a sequence $x_{1}, x_{2}, \ldots, x_{n}$ of real numbers, we define its price as

$$
\max _{1 \leqslant i \leqslant n}\left|x_{1}+\cdots+x_{i}\right|
$$

Given $n$ real numbers, Dave and George want to arrange them into a sequence with a low price. Diligent Dave checks all possible ways and finds the minimum possible price $D$. Greedy George, on the other hand, chooses $x_{1}$ such that $\left|x_{1}\right|$ is as small as possible; among the remaining numbers, he chooses $x_{2}$ such that $\left|x_{1}+x_{2}\right|$ is as small as possible, and so on. Thus, in the $i^{\text {th }}$ step he chooses $x_{i}$ among the remaining numbers so as to minimise the value of $\left|x_{1}+x_{2}+\cdots+x_{i}\right|$. In each step, if several numbers provide the same value, George chooses one at random. Finally he gets a sequence with price $G$.

Find the least possible constant $c$ such that for every positive integer $n$, for every collection of $n$ real numbers, and for every possible sequence that George might obtain, the resulting values satisfy the inequality $G \leqslant c D$.

Solution:
If the initial numbers are $1,-1,2$, and -2 , then Dave may arrange them as $1,-2,2,-1$, while George may get the sequence $1,-1,2,-2$, resulting in $D=1$ and $G=2$. So we obtain $c \geqslant 2$.

Therefore, it remains to prove that $G \leqslant 2 D$. Let $x_{1}, x_{2}, \ldots, x_{n}$ be the numbers Dave and George have at their disposal. Assume that Dave and George arrange them into sequences $d_{1}, d_{2}, \ldots, d_{n}$ and $g_{1}, g_{2}, \ldots, g_{n}$, respectively. Put

$$
M=\max _{1 \leqslant i \leqslant n}\left|x_{i}\right|, \quad S=\left|x_{1}+\cdots+x_{n}\right|, \quad \text { and } \quad N=\max \{M, S\}
$$

We claim that

$$
D \geqslant S,
\tag{1}
$$
$$
D \geqslant \frac{M}{2}, \quad \text { and } 
\tag{2}
$$
$$
G \leqslant N=\max \{M, S\} 
\tag{3}
$$

These inequalities yield the desired estimate, as $G \leqslant \max \{M, S\} \leqslant \max \{M, 2 S\} \leqslant 2 D$.

The inequality (1) is a direct consequence of the definition of the price.

To prove (2), consider an index $i$ with $\left|d_{i}\right|=M$. Then we have

$$
M=\left|d_{i}\right|=\left|\left(d_{1}+\cdots+d_{i}\right)-\left(d_{1}+\cdots+d_{i-1}\right)\right| \leqslant\left|d_{1}+\cdots+d_{i}\right|+\left|d_{1}+\cdots+d_{i-1}\right| \leqslant 2 D
$$

as required.

It remains to establish (3). Put $h_{i}=g_{1}+g_{2}+\cdots+g_{i}$. We will prove by induction on $i$ that $\left|h_{i}\right| \leqslant N$. The base case $i=1$ holds, since $\left|h_{1}\right|=\left|g_{1}\right| \leqslant M \leqslant N$. Notice also that $\left|h_{n}\right|=S \leqslant N$.

For the induction step, assume that $\left|h_{i-1}\right| \leqslant N$. We distinguish two cases.

Case 1. Assume that no two of the numbers $g_{i}, g_{i+1}, \ldots, g_{n}$ have opposite signs.

Without loss of generality, we may assume that they are all nonnegative. Then one has $h_{i-1} \leqslant h_{i} \leqslant \cdots \leqslant h_{n}$, thus

$$
\left|h_{i}\right| \leqslant \max \left\{\left|h_{i-1}\right|,\left|h_{n}\right|\right\} \leqslant N
$$

Case 2. Among the numbers $g_{i}, g_{i+1}, \ldots, g_{n}$ there are positive and negative ones.



Then there exists some index $j \geqslant i$ such that $h_{i-1} g_{j} \leqslant 0$. By the definition of George's sequence we have

$$
\left|h_{i}\right|=\left|h_{i-1}+g_{i}\right| \leqslant\left|h_{i-1}+g_{j}\right| \leqslant \max \left\{\left|h_{i-1}\right|,\left|g_{j}\right|\right\} \leqslant N
$$

Thus, the induction step is established.

Final answer: 2