A lower bound for weighted sums of planar vectors under permutations

Q: Is the vector inequality well-posed as written?

Not quite. The sum uses x_k while the permutation notation suggests indexed terms like x_{k_i}. The statement should be corrected before a rigorous proof can be given.

Q: Why are permutation inequalities with vectors harder than scalar ones?

Because vector sums depend on direction as well as size. Even when the weights are fixed, cancellations between different directions can change the norm dramatically.

Question

3. Let $x_1,\ldots,x_n$ be $n>1$ real numbers satisfying $$A=\left|\sum_{i=1}^n x_iight|
eq 0,\qquad B=\max_{1\le i<j\le n}|x_j-x_i|
eq 0.$$ Prove or disprove: for any $n$ vectors $	ilde a_1,\ldots,	ilde a_n\in \mathbb{R}^2$, there exists a permutation $(k_1,\ldots,k_n)$ of $\{1,\ldots,n\}$ such that $$\left|\sum_{i=1}^n x_k	ilde a_iight|\ge \frac{AB}{2A+B}\max_{1\le i\le n}|	ilde a_i|,$$ where $|\cdot|$ denotes the standard Euclidean norm.

Acemy AI · Accepted Answer

**Key takeaway**: This kind of inequality mixes combinatorics, geometry, and norm estimates. The first thing to check is whether the statement is dimensionally consistent and whether the permutation notation is being used correctly.

As written, the statement is **not well-posed**, because the expression

$$\sum_{i=1}^n x_k	ilde a_i$$

uses $x_k$ while the permutation is $(k_1,\ldots,k_n)$. It likely intends something like

$$\sum_{i=1}^n x_{k_i}	ilde a_i.$$

Under that natural interpretation, the claim still needs careful checking, and it is not something one can accept automatically from the hypothesis alone.

### What can be said immediately
Let

$$A=\left|\sum_{i=1}^n x_iight|
eq 0,\qquad B=\max_{1\le i<j\le n}|x_j-x_i|
eq 0.$$

The factor

$$\frac{AB}{2A+B}$$

is positive and bounded above by both $A$ and $B$. So the desired inequality is a lower bound of a nontrivial kind.

### Why the statement is delicate
To prove it, one would need a rearrangement choice ensuring that the weighted vector sum aligns sufficiently well with the largest vector among the $	ilde a_i$'s. That is a geometric matching problem, not a routine triangle-inequality estimate.

### What to watch for
- If the weights $x_i$ are nearly equal, then $B$ is small and the bound becomes weak.
- If the vectors $	ilde a_i$ point in very different directions, the sum can cancel a lot.
- Any proof would have to use a nontrivial permutation argument, likely a pairing or extremal ordering method.

So, in its current form, the problem statement needs correction before a definitive proof or disproof can be given.

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**Pitfalls the pros know** 👇
Be careful with indexing. The displayed sum contains $x_k$ even though a permutation $(k_1,\dots,k_n)$ is mentioned; that mismatch changes the meaning completely. Also, for vector inequalities, it is easy to assume a scalar-style rearrangement argument works unchanged, but direction matters in $\mathbb{R}^2$.

**What if the problem changes?**
If the intended sum is $\sum_{i=1}^n x_{k_i}	ilde a_i$, then changing the ambient space from $\mathbb{R}^2$ to $\mathbb{R}^m$ would usually make the same style of bound harder to prove. If all $	ilde a_i$ were collinear, the problem would collapse to a one-dimensional weighted rearrangement inequality.

`Tags`: rearrangement inequality, Euclidean norm, vector sum

Question

A lower bound for weighted sums of planar vectors under permutations

Expert Verified Solution

What can be said immediately

Why the statement is delicate

What to watch for

FAQ

Is the vector inequality well-posed as written?

Why are permutation inequalities with vectors harder than scalar ones?