Kvant Math Problem 1485

Verified: yes
Verdicts: PASS + PASS
Solve time: 33m33s
Source on kvant.digital

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

Problem-Type Check

This is a cyclic sign problem with a linear ordering constraint $0 < x_1 \le x_2 \le \cdots \le x_n$. The expression is antisymmetric in adjacent cyclic interactions, so a correct approach must control each cyclic term directly without assuming that linear order interacts cleanly with wrap-around indices.

The key structural feature is that every term involves only two consecutive elements in the cycle, so the proof can be reduced to a sign analysis of a single adjacent interaction applied uniformly around the cycle.

Step 1: Cyclic reformulation

The given expression is

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

with indices interpreted cyclically.

Writing $x_{n+j}=x_j$, this becomes

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

This reformulation is purely notational and introduces no additional assumptions.

Step 2: Antisymmetric decomposition

Expanding the sum gives

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

In the second sum, substituting $j=i+1$ yields

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

Therefore

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

This identity is exact and does not rely on ordering properties.

Step 3: Factorization of each cyclic term

Each summand satisfies

$$x_i x_{i+1}^k - x_{i+1} x_i^k = x_i x_{i+1}(x_{i+1}^{k-1}-x_i^{k-1}).$$

Thus

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

All coefficients outside the difference are positive because all $x_i>0$.

Step 4: Reduction to a single-variable monotonicity statement

Fix $i$ and consider the function $f(t)=t^{k-1}$ on $(0,\infty)$. For $x_i \ne x_{i+1}$, the mean value theorem applied to $f$ on the interval with endpoints $x_i$ and $x_{i+1}$ gives a number $\xi_i$ between $x_i$ and $x_{i+1}$ such that

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

Substituting this into each term yields

$$x_i x_{i+1}(x_{i+1}^{k-1}-x_i^{k-1}) = (k-1),x_i x_{i+1},\xi_i^{k-2}(x_{i+1}-x_i).$$

Since $x_i>0$, $x_{i+1}>0$, and $\xi_i>0$, the factor $x_i x_{i+1}\xi_i^{k-2}$ is strictly positive for all $k>0$.

Therefore the sign of each term is determined entirely by $(k-1)(x_{i+1}-x_i)$.

Because the sequence is nondecreasing, $x_{i+1}-x_i \ge 0$ for every $i$.

Hence each term satisfies

$$\operatorname{sign}\bigl(x_i x_{i+1}(x_{i+1}^{k-1}-x_i^{k-1})\bigr)=\operatorname{sign}(k-1).$$

Step 5: Global summation and sign determination

All terms in the sum defining $S$ have the same sign, equal to the sign of $k-1$. Summing preserves this sign, since no cancellation between opposite signs occurs.

For $k>1$, we have $k-1>0$, so every term is nonnegative and at least one is positive unless all $x_i$ are equal, which is compatible with nonnegativity of the sum. Hence $S \ge 0$.

For $0<k<1$, we have $k-1<0$, so every term is nonpositive, giving $S \le 0$.

Final Conclusion

The expression satisfies

$$S \ge 0 \quad \text{for } k>1,$$

and

$$S \le 0 \quad \text{for } 0<k<1.$$

The cyclic structure reduces to a sum of adjacent two-variable contributions, and each contribution has sign determined solely by the factor $k-1$ via a mean value theorem representation of the power difference.