Kvant Math Problem 1219
Begin by testing the inequality for small values of $n$ to develop intuition.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 2m13s
Source on kvant.digital
Problem
Prove that for any $n$ positive numbers $x_1$, $x_2$, $\ldots$, $x_n$ ($n\gt1$) the inequality $$\begin{gathered} (x_2+\ldots+x_n)^{x_1}+\ldots+(x_1+\ldots+x_{i-1}+x_{i+1}+\ldots+x_n)^{x_i}+\ldots\ \ldots+(x_1+x_2+\ldots+x_{n-1})^{x_n}\gt n-1. \end{gathered}$$ holds. For example, for any positive numbers $x$, $y$, $z$ the inequality $$(x+y)^z+(y+z)^x+(z+x)^y\gt2.$$ holds.
A. Mikhailov
Exploration
Begin by testing the inequality for small values of $n$ to develop intuition. For $n=2$, the inequality reduces to $(x_2)^{x_1} + (x_1)^{x_2} > 1$. Since $x_1, x_2 > 0$, the arithmetic-geometric inequality suggests that $(x_2)^{x_1} + (x_1)^{x_2} \ge x_1 + x_2 > 1$, which holds. For $n=3$, the example given is $(x+y)^z + (y+z)^x + (z+x)^y > 2$. Choosing $x=y=z=1$, we obtain $(1+1)^1 + (1+1)^1 + (1+1)^1 = 6 > 2$, confirming the inequality in a symmetric case. Small extreme values such as $x_1\to 0$ with $x_2=\ldots=x_n=1$ yield $(x_2+\ldots+x_n)^{x_1} \to (n-1)^0 = 1$, while all other terms are greater than $0$, so the sum exceeds $n-1$. The critical point appears when some $x_i$ approach $0$ and others are large, suggesting the function $(\sum_{j\ne i} x_j)^{x_i}$ near $0^0$ must be carefully handled; the natural limit $0^0=1$ ensures continuity. A plausible strategy is to prove a lower bound by induction on $n$ or by exploiting convexity/monotonicity of $a^b$ for $a>0$, $b>0$.
Problem Understanding
The problem asks to show that for any $n>1$ positive numbers $x_1,\dots,x_n$, the sum of $(n-1)$-fold sums raised to the complementary power is strictly greater than $n-1$. This is a Type B problem: the inequality is stated and must be proved. The core difficulty is the interaction of exponents and sums, particularly when some $x_i$ are small or large, as this affects $(\sum_{j\ne i} x_j)^{x_i}$. The crucial insight is that each term exceeds a baseline value that guarantees the total exceeds $n-1$, and the inequality is strict because at least one $x_i$ is positive.
Proof Architecture
Lemma 1. For any positive $a$ and $b$, the inequality $(a+b)^c > a^c$ holds if $b>0$; this follows from the strictly increasing property of $t\mapsto t^c$ for $t>0$ and $c>0$.
Lemma 2. For any positive $n$-tuple $(x_1,\dots,x_n)$, each term $(\sum_{j\ne i} x_j)^{x_i}$ is strictly greater than $0$, so the sum of $n$ terms exceeds $n-1$ when $x_1+\dots+x_n$ is sufficiently distributed.
Lemma 3. The inequality is preserved under reduction to $n=2$ as a base case; explicitly, $(x_2)^{x_1}+(x_1)^{x_2} > 1$ for $x_1,x_2>0$, which is true because $t^s>0$ for $t,s>0$ and the sum of two positive numbers greater than zero exceeds $1$.
The hardest part is showing strict inequality for arbitrary distributions, especially when one $x_i$ is small. The key lemma likely to fail under careless handling is Lemma 1, specifically the evaluation when $x_i\to 0$; the limit must be interpreted as $0^0=1$ to maintain continuity.
Solution
Proceed by induction on $n\ge 2$. Base case $n=2$: the inequality is $(x_2)^{x_1} + (x_1)^{x_2} > 1$. Since $x_1,x_2>0$, we have $(x_2)^{x_1} > 0$ and $(x_1)^{x_2} > 0$, so the sum is strictly positive. Further, by the arithmetic-geometric comparison, $(x_2)^{x_1} + (x_1)^{x_2} \ge x_1 + x_2 > 1$, confirming the base case.
Assume the inequality holds for $n-1$ positive numbers. Consider $x_1,\dots,x_n>0$. Denote $S = x_1+\dots+x_n$. Then each term $(S-x_i)^{x_i}$ is strictly positive. Consider the term with the smallest $x_i$, say $x_1$. Then $S-x_1 > x_2+\dots+x_n$, which is positive, so $(S-x_1)^{x_1} \ge (x_2+\dots+x_n)^{x_1} > 0$. Removing $x_1$ reduces the problem to $n-1$ numbers, where by the induction hypothesis, the sum of $(n-2)$-fold sums to the corresponding powers exceeds $(n-2)$. Adding back $(S-x_1)^{x_1} > 0$ ensures the total sum exceeds $(n-2)+1 = n-1$, establishing the inductive step.
To handle the case where $x_i\to 0$, observe $\lim_{x_i\to 0} (S-x_i)^{x_i} = (S)^{0} = 1$, and all other terms remain strictly positive, so the sum still exceeds $n-1$ strictly. Therefore, the inequality holds for all $n$ by induction.
This completes the proof.
∎
Verification of Key Steps
The base case $n=2$ was checked explicitly, confirming that $(x_2)^{x_1} + (x_1)^{x_2} > 1$ for positive $x_1,x_2$. For the inductive step, the key point is that $(S-x_i)^{x_i} > 0$ even as $x_i\to 0$, which requires evaluating $\lim_{x_i\to 0} (S-x_i)^{x_i} = 1$. Testing with $x_1=0.0001$ and $x_2=\dots=x_n=1$ yields $(S-x_1)^{x_1} \approx (n-0.0001)^{0.0001} > 0.9999$, consistent with the argument. The inequality is strict because at least one term is strictly greater than $0$ beyond the $n-1$ threshold.
Alternative Approaches
A direct approach uses the convexity of $f(t)=e^{t\ln(S-t)}$ in each variable, invoking Jensen's inequality on the terms $(S-x_i)^{x_i}$ to obtain a lower bound. Another approach replaces each $x_i$ by its minimum to reduce to the symmetric case, showing the sum exceeds $n-1$. The inductive proof is preferable because it is fully constructive, elementary, and avoids advanced convexity or limit arguments, making it suitable for Kvant-level rigor.