Kvant Math Problem 1177

Consider the inequality for small values of $n$ to understand its structure.

kvantmathematicsolympiad

Verified: no
Verdicts: SKIP + SKIP
Solve time: 7m37s
Source on kvant.digital

Problem

Prove that for positive numbers $x_1$, $x_2$, $\ldots$, $x_n$, not exceeding 1, the inequality $$(1+x_1)^{1/x_2}(1+x_2)^{1/x_3}(1+x_3)^{1/x_4}\ldots(1+x_n)^{1/x_1}\ge2^n.$$ holds.

K. P. Kokhas, V. M. Televka

Exploration

Consider the inequality for small values of $n$ to understand its structure. For $n=2$, the inequality reads

$(1+x_1)^{1/x_2}(1+x_2)^{1/x_1}\ge 4.$

Trying $x_1=x_2=1$ gives equality: $(1+1)^{1}(1+1)^1=4$. Choosing $x_1=1$, $x_2=\frac{1}{2}$ gives $(2)^{2.0}(1.5)^{1}=4\cdot 1.5=6>4$, suggesting the left-hand side increases if the numbers are unequal but positive and at most 1. For $n=3$, setting all $x_i=1$ gives $(2)^1(2)^1(2)^1=8=2^3$, again equality. Choosing one $x_i$ less than 1 and the others 1 yields a product greater than $2^3$, suggesting equality occurs when all $x_i=1$.

The cyclic nature of the exponents hints that standard inequalities like AM–GM or convexity may be useful. Observing $(1+x)^{1/x}$ on $(0,1]$, its derivative can be analyzed; preliminary calculations suggest it decreases on $(0,1]$. This indicates that replacing all $x_i$ by the maximal allowed value 1 could minimize the product, yielding $2^n$.

The critical point likely involves showing that for $0<x\le1$, $(1+x)^{1/x}\ge 2$, with equality only at $x=1$. Once this is established, the inequality follows by monotonicity and the cyclic structure.

Problem Understanding

The problem asks to prove an inequality for positive numbers $x_1,\dots,x_n$ bounded above by 1, in a cyclic exponential product. The problem type is B, a pure proof, since the inequality is given and must be established for all admissible $x_i$. The core difficulty is controlling the cyclic exponents and recognizing the minimal configuration. Preliminary exploration suggests the product is minimized when all $x_i=1$. Therefore, the key lemma is that $(1+x)^{1/x}\ge 2$ for $0<x\le1$, which allows the entire cyclic product to be bounded below by $2^n$.

Proof Architecture

Lemma 1: For $0<x\le1$, $(1+x)^{1/x}\ge 2$, with equality if and only if $x=1$. This is true because the function $f(x)=\ln(1+x)/x$ decreases on $(0,1]$, which can be verified by computing $f'(x) = \frac{1}{x(1+x)} - \frac{\ln(1+x)}{x^2}$ and showing it is negative.

Lemma 2: For positive numbers $x_1,\dots,x_n\le 1$, the product $(1+x_1)^{1/x_2}\dots(1+x_n)^{1/x_1}$ is bounded below by replacing each $x_i$ in the base by 1 and each exponent by the corresponding $1$, yielding $\prod_{i=1}^{n}(1)^{1}$.

The main argument applies Lemma 1 to each term, using the cyclic permutation of exponents to show that every factor is at least 2, and then multiplying yields $2^n$. The hardest step is verifying Lemma 1 rigorously, as a careless derivative computation could produce an incorrect monotonicity conclusion.

Solution

Define the function $f(x)=\frac{\ln(1+x)}{x}$ for $0<x\le1$. Its derivative is

$$f'(x)=\frac{1}{x(1+x)}-\frac{\ln(1+x)}{x^2} = \frac{x - (1+x)\ln(1+x)}{x^2(1+x)}.$$

Denote $g(x) = x - (1+x)\ln(1+x)$. Then $g'(x)=1 - \ln(1+x) - \frac{1+x}{1+x} = -\ln(1+x)$, which is negative for $0<x\le1$. Therefore, $g(x)$ decreases on $(0,1]$, and $g(0^+)=0$. Hence $g(x)<0$ for all $0<x\le1$, implying $f'(x)<0$ and so $f(x)$ is strictly decreasing on $(0,1]$. Consequently, for $0<x\le1$, $f(x)\ge f(1)=\ln2$, or equivalently $(1+x)^{1/x} \ge 2$, with equality only at $x=1$.

Applying this inequality to each factor of the cyclic product, we have $(1+x_i)^{1/x_{i+1}}\ge 2$ for $i=1,\dots,n$, with indices taken modulo $n$. Multiplying all $n$ inequalities gives

$$(1+x_1)^{1/x_2}(1+x_2)^{1/x_3}\dots(1+x_n)^{1/x_1} \ge 2^n.$$

Equality occurs only if all $x_i=1$, which simultaneously satisfies each individual equality.

This completes the proof.

∎

Verification of Key Steps

To confirm that $f(x) = \ln(1+x)/x$ decreases, compute $g(x)=x-(1+x)\ln(1+x)$ at endpoints: $g(0^+)=0$, $g(1)=1-2\ln2\approx -0.386<0$. The derivative $g'(x)=-\ln(1+x)<0$ for $0<x\le1$, confirming strict decrease. Testing several values $x=1/2$, $g(1/2)\approx -0.096<0$ validates the negativity numerically.

For the cyclic product, small examples $n=2$, $x_1=1$, $x_2=1/2$ yield $(2)^2(3/2)^1=6>4$, verifying that inequality is maintained and that equality occurs only when all $x_i=1$.

The product of minimal factors is correctly $2^n$, as each term independently attains its minimum at $x_i=1$, ensuring the final bound is sharp.

Alternative Approaches

An alternative proof uses the logarithm and AM–GM inequality. Taking logarithms, the cyclic sum $\sum_{i=1}^{n} \frac{\ln(1+x_i)}{x_{i+1}}$ can be bounded using convexity of $\ln(1+x)$ and the decreasing property of $f(x)$. This method ultimately reduces to the same key inequality $(1+x)^{1/x}\ge2$, but involves more manipulations and additional inequalities, making the derivative approach more direct and transparent. Another approach could involve induction on $n$, proving the statement for $n-1$ and adding the $n$-th term, but this introduces unnecessary complications. The main approach is preferable because it isolates the critical function and minimizes auxiliary arguments.