Problem A. 921. (December 2025)

A. 921. Let \(\displaystyle n \ge 2\) be an integer. Let \(\displaystyle x_1,\ldots,x_n\) be positive real numbers whose product is \(\displaystyle n-1\). Prove that

\(\displaystyle x_1+\ldots+x_n-\left(\frac{1}{x_1^{n-1}}+\ldots+\frac{1}{x_n^{n-1}}\right)<(n-1)^{1+\frac{1}{n}}.\)

Proposed by Áron Bán-Szabó, Palaiseau

(7 pont)

Deadline expired on January 12, 2026.

Let us rewrite the variables in the form \(\displaystyle x_i=e^{t_i}\). This is possible since all \(\displaystyle x_i\) are positive. The condition on the product then becomes \(\displaystyle t_1+t_2+\dots+t_n=\log_e(n-1)\), so the sum \(\displaystyle t_1+\dots+t_n\) is fixed. The quantity to be maximized can be written as \(\displaystyle f(t_1)+f(t_2)+\dots+f(t_n)\), where \(\displaystyle f(t)=e^t-e^{-(n-1)t}\).

We now study the convexity of \(\displaystyle f\). We have \(\displaystyle f'(t)=e^t+(n-1)e^{-(n-1)t}\) and \(\displaystyle f''(t)=e^t-(n-1)^2e^{-(n-1)t}\). The equality \(\displaystyle f''(t)=0\) holds at \(\displaystyle t=\frac{\log_e((n-1)^2)}{n}\). Moreover, \(\displaystyle f''\) is increasing since \(\displaystyle f'''(t)=e^t+(n-1)^3e^{-(n-1)t}>0\). Hence \(\displaystyle f\) is concave for \(\displaystyle t<\frac{\log_e((n-1)^2)}{n}\) and convex for \(\displaystyle t>\frac{\log_e((n-1)^2)}{n}\).

If we pick two variables \(\displaystyle t_i\) and \(\displaystyle t_j\) that lie in the convex region of \(\displaystyle f\), then the sum \(\displaystyle f(t_i)+f(t_j)\) increases when we move them further apart while keeping \(\displaystyle t_i+t_j\) fixed. In contrast, if two variables lie in the concave region, then \(\displaystyle f(t_i)+f(t_j)\) increases when we move them closer to each other while keeping \(\displaystyle t_i+t_j\) fixed. Using the first type of step, we can arrange that there is at most one variable in the convex region. Using the second type of step, we can arrange that all variables in the concave region are equal to each other. Indeed, among the variables in the concave region, if they are not all equal, then there exists one that is larger than their average and one that is smaller than their average; we move these two closer to each other until one of them becomes equal to the average. Repeating this procedure, we conclude that in searching for the maximum of \(\displaystyle f(t_1)+\dots+f(t_n)\) we may assume that at least \(\displaystyle n-1\) variables are equal.

Thus we may assume \(\displaystyle x_1=x_2=\dots=x_{n-1}=x\). Then the constraint gives \(\displaystyle x^{n-1}x_n=n-1\), i.e. \(\displaystyle x_n=\frac{n-1}{x^{n-1}}\). Therefore it suffices to consider the following one-variable function:

\(\displaystyle g(x)=(n-1)x+\frac{n-1}{x^{n-1}}-(n-1)\frac{1}{x^{n-1}}-\frac{x^{(n-1)^2}}{(n-1)^{n-1}} =(n-1)x-\frac{x^{(n-1)^2}}{(n-1)^{n-1}}. \)

Differentiating,

\(\displaystyle g'(x)=(n-1)-\frac{(n-1)^2x^{n^2-2n}}{(n-1)^{n-1}}. \)

Solving \(\displaystyle g'(x)=0\) over the positive real numbers yields the unique solution \(\displaystyle x=(n-1)^{1/n}\). Since \(\displaystyle g'\) is strictly decreasing on \(\displaystyle (0,\infty)\), the function \(\displaystyle g\) is concave on the positive reals, and hence \(\displaystyle g\) attains its maximum at \(\displaystyle x=(n-1)^{1/n}\). The value at this point is less than \(\displaystyle (n-1)x=(n-1)\cdot (n-1)^{1/n}=(n-1)^{1+1/n}\), which completes the proof.

Statistics:

19 students sent a solution.
7 points:	Ali Richárd, Aravin Peter, Bodor Mátyás, Bolla Donát Andor, Diaconescu Tashi, Forrai Boldizsár, Gyenes Károly, Rajtik Sándor Barnabás, Vigh 279 Zalán, Xiaoyi Mo.
6 points:	Sárdinecz Dóra, Szakács Ábel.
5 points:	1 student.
2 points:	2 students.
1 point:	2 students.
0 point:	2 students.

Problems in Mathematics of KöMaL, December 2025