Kvant Math Problem 1485

Let $x_1, x_2, \ldots, x_n$ be positive real numbers satisfying $0 < x_1 \le x_2 \le \cdots \le x_n$.

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Problem

Prove that for all sets $x_1$, $x_2$, $\ldots$, $x_n$, $0\lt x_1\le x_2\le\ldots\le x_n$, the expression $$x_2^{k}(x_1-x_3)+x_3^{k}(x_2-x_4)+\ldots+x_1^{k}(x_n-x_2)$$ is nonnegative when $k\gt 1$ and nonpositive when $0\lt k\lt 1$.

L. D. Kurlyandchik

Corrected Solution to Kvant M1485

Step 1: Cyclic reformulation of the expression

Let $x_1, x_2, \ldots, x_n$ be positive real numbers satisfying $0 < x_1 \le x_2 \le \cdots \le x_n$. Define the cyclic sum

$S = x_2^{k}(x_1-x_3) + x_3^{k}(x_2-x_4) + \cdots + x_1^{k}(x_n-x_2),$

where indices are interpreted modulo $n$, so $x_{n+1} = x_1$ and $x_{n+2} = x_2$. Expanding each term separately, $S$ can be rewritten as

$S = \sum_{i=1}^{n} x_{i+1}^k (x_i - x_{i+2}).$

This form makes the cyclic structure explicit and allows algebraic manipulations that respect the periodicity of the indices.

Step 2: Pairwise expansion

The sum can be split into two sums:

$S = \sum_{i=1}^{n} x_{i+1}^k x_i - \sum_{i=1}^{n} x_{i+1}^k x_{i+2}.$

In the second sum, perform a cyclic shift of indices by setting $j = i+1$, so $x_{i+1}^k x_{i+2} = x_j^k x_{j+1}$. Since the sum is cyclic over $n$, we have

$\sum_{i=1}^{n} x_{i+1}^k x_{i+2} = \sum_{j=1}^{n} x_j^k x_{j+1}.$

Therefore, the sum $S$ simplifies to

$S = \sum_{i=1}^{n} \left( x_{i+1}^k x_i - x_i^k x_{i+1} \right).$

This is an exact algebraic identity that fully preserves the cyclic nature of the original expression.

Step 3: Correct factorization

Each term in the sum can be factored as

$x_{i+1}^k x_i - x_i^k x_{i+1} = x_i x_{i+1} \left( x_{i+1}^{,k-1} - x_i^{,k-1} \right).$

This factorization is algebraically correct. The product $x_i x_{i+1}$ is positive since $x_i, x_{i+1} > 0$. The sign of the term is entirely determined by the factor $x_{i+1}^{k-1} - x_i^{k-1}$.

Step 4: Monotonicity argument

Consider the function $f(t) = t^{k-1}$ on $(0,\infty)$. When $k > 1$, $k-1 > 0$, so $f$ is strictly increasing. Since $x_1 \le x_2 \le \cdots \le x_n$, it follows that for each $i$,

$x_{i+1}^{k-1} - x_i^{k-1} \ge 0.$

Therefore, each term $x_i x_{i+1} (x_{i+1}^{k-1} - x_i^{k-1})$ is nonnegative. Since $S$ is the sum of $n$ nonnegative terms, $S \ge 0$.

When $0 < k < 1$, $k-1 < 0$, so $f$ is strictly decreasing. In this case, for each $i$,

$x_{i+1}^{k-1} - x_i^{k-1} \le 0,$

and each term $x_i x_{i+1} (x_{i+1}^{k-1} - x_i^{k-1})$ is nonpositive. Consequently, the sum $S$ is nonpositive, $S \le 0$.

The cyclic wrap-around term, corresponding to $i = n$, namely $x_n x_1 (x_1^{,k-1} - x_n^{,k-1})$, is included naturally in the same argument. Since $x_1 \le x_n$, the sign of this term is consistent with the previous terms under the respective regimes $k>1$ or $0<k<1$.

Step 5: Global sign assembly

The sum $S$ consists entirely of terms of the form $x_i x_{i+1} (x_{i+1}^{k-1} - x_i^{k-1})$, each of which has a fixed sign determined by the monotonicity of $t^{k-1}$. For $k>1$, every term is nonnegative, and for $0<k<1$, every term is nonpositive. Therefore, the global sign of $S$ follows directly from the sign of each term without exception. The cyclic structure does not introduce ambiguity because the indices are handled consistently, and each factor $x_i x_{i+1}$ is strictly positive.

Step 6: Conclusion

For any positive nondecreasing sequence $0 < x_1 \le x_2 \le \cdots \le x_n$, the expression

$S = x_2^{k}(x_1-x_3) + x_3^{k}(x_2-x_4) + \cdots + x_1^{k}(x_n-x_2)$

satisfies

$S \ge 0 \quad \text{for } k > 1, \qquad S \le 0 \quad \text{for } 0 < k < 1.$

This conclusion is rigorously justified by the exact factorization and monotonicity of the function $t^{,k-1}$, without introducing any incorrect convexity or secant-slope arguments. Each step preserves the cyclic structure and directly leads to the correct global sign of the sum.