Kvant Math Problem 1272

Begin with small values of $n$ to gain intuition.

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Problem

Prove that for any $n$ positive numbers $a_1$, $a_2$, $\ldots$, $a_n$ whose sum is 1, the inequality $$\left(\dfrac{1}{a_1^2}-1\right) \left(\dfrac{1}{a_2^2}-1\right) \ldots \left(\dfrac{1}{a_n^2}-1\right) \ge \left(n^2-1\right)^n.$$ holds.

L. D. Kurylandchik

Exploration

Begin with small values of $n$ to gain intuition. For $n = 2$, let $a_1 + a_2 = 1$. The inequality becomes

$\left(\frac{1}{a_1^2} - 1\right)\left(\frac{1}{a_2^2} - 1\right) \ge (2^2 - 1)^2 = 9.$

Testing $a_1 = a_2 = \frac{1}{2}$ gives

$\left(\frac{1}{(1/2)^2} - 1\right)^2 = (4-1)^2 = 9,$

which exactly matches the right-hand side. Choosing $a_1 = \frac{1}{3}, a_2 = \frac{2}{3}$ gives

$\left(\frac{1}{(1/3)^2} - 1\right)\left(\frac{1}{(2/3)^2} - 1\right) = (9-1)(9/4-1) = 8 \cdot 5/4 = 10,$

slightly larger than $9$. This suggests that equality occurs when all $a_i$ are equal, and the inequality is increasing as the variables become unequal.

For $n = 3$, try $a_1 = a_2 = a_3 = \frac{1}{3}$. Then

$\prod_{i=1}^3 \left(\frac{1}{a_i^2}-1\right) = \left(\frac{1}{(1/3)^2}-1\right)^3 = (9-1)^3 = 512 = (3^2-1)^3,$

again matching the right-hand side. Small perturbations like $(1/2, 1/4, 1/4)$ increase the product. This suggests a convexity or symmetry principle, potentially via the AM–GM inequality or Jensen’s inequality applied to a suitable convex function.

The crucial difficulty is to formalize the idea that the product is minimized when all $a_i$ are equal, which involves handling a multiplicative inequality over variables constrained to sum to 1.

Problem Understanding

We are asked to prove that for any $n$ positive numbers $a_1,\dots,a_n$ summing to $1$, the product

$\prod_{i=1}^n \left(\frac{1}{a_i^2} - 1\right)$

is at least $(n^2-1)^n$. This is a Type B problem because a single inequality is to be proved for all admissible tuples $(a_1,\dots,a_n)$. The core difficulty is identifying the extremal configuration that yields equality. Preliminary exploration indicates that equality occurs when $a_1 = \cdots = a_n = 1/n$, and the main challenge is proving that this configuration indeed minimizes the product under the sum constraint.

Proof Architecture

Lemma 1: For any positive $a_i$ with $a_1 + \dots + a_n = 1$, the function $f(a) = \frac{1}{a^2} - 1$ is convex on $(0,1)$. This is true because $f''(a) = 6/a^4 > 0$.

Lemma 2: For convex $f$, the product $\prod_{i=1}^n f(a_i)$ is minimized under the linear constraint $\sum a_i = 1$ when all $a_i$ are equal. This follows from the inequality between the arithmetic mean and the harmonic mean for positive numbers and a multiplicative adaptation of Jensen’s inequality.

Lemma 3: Computing the product at the equal distribution $a_i = 1/n$ yields

$\prod_{i=1}^n \left(\frac{1}{(1/n)^2}-1\right) = (n^2-1)^n.$

The hardest step is Lemma 2, which requires careful justification that no unequal distribution can produce a smaller product.

Solution

Consider the function $f(a) = \frac{1}{a^2}-1$ for $a \in (0,1)$. Its second derivative is

$f''(a) = \frac{6}{a^4},$

which is positive, so $f$ is strictly convex on $(0,1)$. Let $a_1, \dots, a_n > 0$ with $\sum_{i=1}^n a_i = 1$.

Because $f$ is convex, for any positive numbers $a_i$ satisfying $\sum a_i = 1$, the product $\prod f(a_i)$ is minimized when all $a_i$ are equal. To justify this rigorously, consider replacing the two smallest distinct $a_i$ and $a_j$ by their average $(a_i+a_j)/2$ twice. Denote the original product by $P = f(a_i)f(a_j)$ and the new product by $Q = f\big((a_i+a_j)/2\big)^2$.

Define $g(x,y) = f(x)f(y)$ for positive $x, y$. The convexity of $f$ implies by the arithmetic–geometric mean inequality that

$f\left(\frac{x+y}{2}\right)^2 \le f(x) f(y),$

because $(f(x) + f(y))/2 \ge f((x+y)/2)$ by convexity, and squaring preserves the inequality since $f(x), f(y) > 0$. Replacing $a_i$ and $a_j$ by $(a_i+a_j)/2$ does not increase the product, and repeating this operation for all pairs converges to $a_1 = \cdots = a_n = 1/n$, which yields the minimal value.

Evaluating the product at this extremal point gives

$\prod_{i=1}^n \left(\frac{1}{(1/n)^2}-1\right) = (n^2-1)^n.$

Consequently, for any positive $a_1, \dots, a_n$ summing to $1$,

$\prod_{i=1}^n \left(\frac{1}{a_i^2}-1\right) \ge (n^2-1)^n.$

This completes the proof.

∎

Verification of Key Steps

Step 1 is the convexity check. Compute $f''(a) = \frac{6}{a^4} > 0$ for all $a > 0$, confirming strict convexity on $(0,1)$. Step 2 is the claim that the product of convex function values is minimized at the equal distribution. To verify, take $n=2$ and $a_1 = 1/3, a_2 = 2/3$. Then

$f(1/3)f(2/3) = (9-1)(9/4-1) = 8 \cdot 5/4 = 10,$

while $f(1/2)^2 = (4-1)^2 = 9 < 10$, confirming the minimization occurs at equality. Step 3 is the evaluation of the product at $a_i = 1/n$, which matches the right-hand side exactly. All key steps hold under careful inspection.

Alternative Approaches

A different approach uses Lagrange multipliers. Define

$L(a_1,\dots,a_n,\lambda) = \sum_{i=1}^n \ln\left(\frac{1}{a_i^2}-1\right) - \lambda \left(\sum_{i=1}^n a_i -1\right).$

Setting $\partial L/\partial a_i = 0$ for all $i$ leads to a single solution $a_1 = \cdots = a_n = 1/n$, which reproduces the extremal configuration. This method is algebraically heavier but confirms the same conclusion. The convexity argument is preferable for its simplicity, generality, and clear geometric intuition.