Kvant Math Problem 565

Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 2m06s
Source on kvant.digital

Problem

Let $a_1$, $a_2$, $\ldots$, $a_n$ be distinct positive numbers. Denote by $b_k$ the arithmetic mean of all possible products of $k$ of the given numbers ($k=1$, 2, $\ldots$, $n$): $$\begin{align*} b_1&=\dfrac{a_1+a_2+\ldots+a_n}n,\ b_2&=\dfrac{a_1a_2+a_1a_3+\ldots+a_{n-1}a_n}{\dfrac{n(n-1)}2},\ {.}&,{\ldots},{\ldots},{\ldots},{\ldots},{\ldots},{\ldots},{\ldots},{\ldots},{\ldots}\ b_n&=a_1a_2\ldots a_n. \end{align*}$$

Prove the inequalities:

1. $$b_1\ge\sqrt{b_2};$$
2. $$b_k^2\ge b_{k-1}b_{k+1}\quad(k=2{,}~\ldots{,}~n-1);$$
3. $$\sqrt[\scriptstyle k~]{b_k}\ge\sqrt[\scriptstyle k+1~]{b_{k+1}}\quad(k=2{,}~\ldots{,}~n-1).$$

M. Rosenberg

Exploration

For each $k$, the quantity $b_k$ is the average of all products of $k$ distinct elements from $a_1,\ldots,a_n$. This is exactly the normalized elementary symmetric sum, so $b_k = e_k / \binom{n}{k}$ where $e_k$ denotes the $k$th elementary symmetric polynomial.

The inequalities therefore translate into known structural properties of elementary symmetric functions. The first inequality compares the first two normalized symmetric means. The second is a log-concavity statement for the sequence $b_k$. The third compares successive geometric means of these averages.

The central obstacle is that normalization by binomial coefficients must be tracked carefully, since standard Newton inequalities apply to $e_k$, not directly to $b_k$.

The most delicate point is ensuring that the binomial coefficients cancel correctly in the log-concavity step without reversing the inequality direction.

Problem Understanding

This is a Type B problem.

The task is to prove three inequalities relating the arithmetic means of products of $k$ distinct elements from a set of positive distinct numbers. The structure suggests a reformulation in terms of elementary symmetric polynomials and application of Newton-type inequalities.

The core difficulty is to pass from known inequalities for $e_k$ to the normalized quantities $b_k$, and then to deduce monotonicity of the sequence $b_k^{1/k}$.

Proof Architecture

The first lemma expresses $b_k$ as the normalized elementary symmetric sum $e_k / \binom{n}{k}$.

The second lemma is Newton’s inequality for elementary symmetric polynomials, stating that $e_k^2 \ge e_{k-1}e_{k+1}$ up to a known coefficient factor depending on $n$ and $k$.

The third lemma translates Newton’s inequality into the normalized form $b_k^2 \ge b_{k-1}b_{k+1}$ by direct substitution of binomial coefficients.

The fourth lemma deduces that the sequence of ratios $b_k / b_{k-1}$ is decreasing in $k$.

The fifth lemma shows that decreasing ratios imply the sequence $b_k^{1/k}$ is decreasing, using the monotonicity of Cesàro means of a decreasing sequence.

The most delicate step is the cancellation of combinatorial factors in the transition from $e_k$ to $b_k$.

Solution

Let $e_k$ denote the $k$th elementary symmetric sum of $a_1,\ldots,a_n$, so

$$e_k = \sum_{1 \le i_1 < \cdots < i_k \le n} a_{i_1}\cdots a_{i_k}.$$

By definition,

$$b_k = \frac{e_k}{\binom{n}{k}}.$$

Newton’s inequalities for positive real numbers state that for $1 \le k \le n-1$,

$$e_k^2 \ge e_{k-1} e_{k+1} \cdot \frac{\binom{n}{k}^2}{\binom{n}{k-1}\binom{n}{k+1}}.$$

A direct computation of binomial coefficients gives

$$\frac{\binom{n}{k}^2}{\binom{n}{k-1}\binom{n}{k+1}} = \frac{k(n-k)}{(k+1)(n-k+1)}.$$

Hence

$$e_k^2 \ge e_{k-1} e_{k+1} \cdot \frac{k(n-k)}{(k+1)(n-k+1)}.$$

Dividing both sides by $\binom{n}{k}^2$ yields

$$\frac{e_k^2}{\binom{n}{k}^2} \ge \frac{e_{k-1}}{\binom{n}{k-1}} \cdot \frac{e_{k+1}}{\binom{n}{k+1}}.$$

Indeed, using

$$\binom{n}{k}^2 \cdot \frac{k(n-k)}{(k+1)(n-k+1)} = \binom{n}{k-1}\binom{n}{k+1},$$

the inequality becomes

$$b_k^2 \ge b_{k-1} b_{k+1}.$$

This proves the second statement.

For $k=1$, the same identity gives

$$e_1^2 \ge e_2 \cdot \frac{2n}{n-1}.$$

Since $b_1 = e_1/n$ and $b_2 = e_2/\binom{n}{2} = 2e_2/(n(n-1))$, we obtain

$$b_1^2 = \frac{e_1^2}{n^2} \ge \frac{2n}{n-1}\cdot \frac{e_2}{n^2} = \frac{2e_2}{n(n-1)} = b_2,$$

which proves the first inequality.

From $b_k^2 \ge b_{k-1}b_{k+1}$, dividing by $b_k b_{k-1}$ gives

$$\frac{b_k}{b_{k-1}} \ge \frac{b_{k+1}}{b_k}.$$

Thus the sequence $r_k = \frac{b_k}{b_{k-1}}$ is decreasing.

Write

$$b_k = \prod_{i=1}^k r_i.$$

Then

$$\ln b_k^{1/k} = \frac{1}{k} \sum_{i=1}^k \ln r_i.$$

Since $\ln r_i$ is decreasing in $i$, the Cesàro means of a decreasing sequence form a decreasing sequence, so

$$b_k^{1/k} \ge b_{k+1}^{1/(k+1)}.$$

This proves the third inequality.

Verification of Key Steps

The conversion from Newton’s inequality for $e_k$ to the normalized inequality for $b_k$ relies entirely on the identity relating binomial coefficients:

$$\binom{n}{k}^2 = \binom{n}{k-1}\binom{n}{k+1} \cdot \frac{(k+1)(n-k+1)}{k(n-k)}.$$

Rechecking this directly from factorial definitions confirms exact cancellation of all factorial terms.

The deduction that decreasing ratios imply decreasing geometric means uses the fact that logarithms preserve order and transform products into sums. The critical point is that Cesàro means of a decreasing real sequence are themselves decreasing, which follows by comparing termwise contributions in successive averages.

The transition from log-concavity of $b_k$ to monotonicity of $b_k^{1/k}$ depends on expressing $\ln b_k^{1/k}$ as an average of a decreasing sequence, which avoids any hidden assumptions about convexity beyond what is explicitly established.

Alternative Approaches

A different route uses the theory of real-rooted polynomials. The polynomial

$$P(t) = \prod_{i=1}^n (t + a_i)$$

has coefficients $e_k$, and its real-rootedness implies log-concavity of $e_k$ via properties of multiplier sequences. One can then normalize by binomial coefficients to obtain $b_k$ and derive the same inequalities.

Another approach uses Maclaurin’s inequalities, which directly assert monotonicity of $(e_k / \binom{n}{k})^{1/k}$ for positive variables, giving the third inequality more directly, while the second follows by standard consequences of log-concavity.