# Is the AM-GM inequality the only obstruction for getting a specific sum and product?

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 $$\lim_{a \to \infty} f(a) = + \infty \, .$$

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 $$S_1 \ge \sqrt{S_2} \ge \sqrt{S_3} \ge \ldots \ge \sqrt[n]{S_n} \, .$$ 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 $$S_1 \ge \sqrt{S_2} \ge \sqrt{S_3} \ge \ldots \ge \sqrt[n]{S_n} \, .$$ 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 $$\frac{a_1+a_2+a_3}{3} = S_1, \, \frac{a_1a_2 + a_1 a_3 + a_2 a_3}{3} = S_2,\, a_1 a_2 a_3 = S_3 \, .$$

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.

• 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. – Misha Lavrov May 3 at 18:49
• @MishaLavrov See edit. – J.G. May 3 at 18:56
• Okay, that's easier than how I wanted to check positivity :) – Misha Lavrov May 3 at 18:59