This might be silly, but here it goes.

Let $P,S>0$ be positive real numbers that satisfy $\frac{S}{n} \ge \sqrt[n]{P}$.

Does there exist a sequence of positive real numbers $a_1,\dots,a_n$ such that $S=\sum a_i,P=\prod a_i$?

Clearly, $\frac{S}{n} \ge \sqrt[n]{P}$ is a necessary condition, due to the AM-GM inequality. But is it sufficient?

For $n=2$, the answer is positive, as can be seen by analysing the discriminant of the associated quadratic equation. (In fact, the solvability criterion for the quadratic, namely- the non-negativity of the discriminant, is equivalent to the AM-GM inequality for the sum and the product).

What about $n>3$?


If we choose $$ (a_1, \ldots, a_{n-1}, a_n) = (a, \ldots, a, \frac{P}{a^{n-1}}) $$ for some $a > 0$ then $\prod a_i = P$ is satisfied, and we need that $$ f(a) = \sum a_i = (n-1)a + \frac{P}{a^{n-1}} = S \, . $$ This equation has a solution because $f$ is continuous on $(0, \infty)$ with $$ \min_{a > 0} f(a) = f(\sqrt[n]P) = n \sqrt[n]P \le S $$ and $$ \lim_{a \to \infty} f(a) = + \infty \, . $$

Remarks about a possible generalization

The Maclaurin's inequalities state the following: If $a_1, \ldots, a_n$ are positive real numbers, and the “averages” $S_1, \ldots, S_n$ are defined as $$ S_k = \frac{1}{\binom n k} \sum_{1 \le i_1 < \ldots < i_k \le n} a_{i_1}a_{i_2} \cdots a_{i_k} $$ then $$ S_1 \ge \sqrt{S_2} \ge \sqrt[3]{S_3} \ge \ldots \ge \sqrt[n]{S_n} \, . $$ In particular, $$ \frac 1 n (a_1 + \ldots + a_n) = S_1 \ge \sqrt[n]{S_n} = \sqrt[n]{a_1 \cdots a_n} $$ so that this is a generalization of the inequality between the geometric and arithmetic means. Therefore a natural generalization of the above question would be:

Let $S_1, \ldots, S_n$ be given positive real numbers with $$ S_1 \ge \sqrt{S_2} \ge \sqrt[3]{S_3} \ge \ldots \ge \sqrt[n]{S_n} \, . $$ Are there positive real numbers $a_1, \ldots, a_n$ such that $$ S_k = \frac{1}{\binom n k} \sum_{1 \le i_1 < \ldots < i_k \le n} a_{i_1}a_{i_2} \cdots a_{i_k} $$ for $1 \le k \le n$?

This is equivalent to asking if the polynomial $$ p(x) = x^n - \binom n 1 S_1 x^{n-1} + \binom n 2 S_2 x_{n-2} + \ldots + (-1)^n \binom n n S_n $$ has $n$ positive real zeros.

Unfortunately, this generalization does not hold. The following counterexample is from

  • SIKLOS, STEPHEN. “Maclaurin's Inequalities: Reflections on a STEP Question.” The Mathematical Gazette, vol. 96, no. 537, 2012, pp. 499–507. JSTOR, https://www.jstor.org/stable/24496873.

We choose $n=3$ and $$ S_1 = \frac 3 2, \, S_2 = 2, \, S_3 = 1 \, . $$ The condition on the $S_k$ is satisfied (with strict inequalities), but a simple analysis shows that the polynomial $$ p(x) = x^3 - 3 S_1 x^2 + 3 S_2 x - S_3 = x^3 - \frac 9 2 x^2 + 6 x - 1 $$ has one real (positive) zero and two non-real zeros. It is therefore not possible to find real numbers $a_1, a_2, a_3$ such that $$ \frac{a_1+a_2+a_3}{3} = S_1, \, \frac{a_1a_2 + a_1 a_3 + a_2 a_3}{3} = S_2,\, a_1 a_2 a_3 = S_3 \, . $$

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Let $\mu:=\frac{S}{n},\,\sigma:=P^{1/n}\le\mu$. We seek $a_i$ of AM $\mu$ & GM $\sigma$. Take $a_1=\cdots=a_{n-2}=\sigma$ so we want $a_{n-1}+a_n=n\mu-(n-2)\sigma=n(\mu-\sigma)+2\sigma,\,a_{n-1}a_n=\sigma^2$. This is achievable because it implies$$(a_{n-1}+a_n)^2-4a_{n-1}a_n=(n(\mu-\sigma)+2\sigma)^2-4\sigma^2\ge0.$$Edit: it's been pointed out we need to verify these roots of a quadratic equation are positive. That's easy: their product is positive, as is their sum.

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  • 2
    $\begingroup$ The advantage of this approach is that it gives us a quadratic equation to solve: $a_{n-1}, a_n$ are the two roots of $x^2 - [n(\mu-\sigma)+2\sigma] x + \sigma^2 = 0$. The other answer requires solving a high-degree equation if we want to find an example of $a_1, \dots, a_n$ that work. $\endgroup$ – Misha Lavrov May 3 at 18:49
  • $\begingroup$ @MishaLavrov See edit. $\endgroup$ – J.G. May 3 at 18:56
  • 1
    $\begingroup$ Okay, that's easier than how I wanted to check positivity :) $\endgroup$ – Misha Lavrov May 3 at 18:59

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