A very elementary problem that I want to see if its solvable or not

Question

Here's the problem:

I wrote this as part of a test I'm writing and I want to see if it's solvable or not.

Answer 1

The AM-GM inequality states that the arithmetic mean is strictly greater than the geometric mean for distinct positive reals. Hence, for $AM-GM$ to be zero we require that all elements in the set be identical, and so no solution is possible.

Edit: From Bollobás' Linear Analysis:

Theorem: The geometric mean of $n$ non-negative reals does not exceed their arithmetic mean: if $a=(a_1,\ldots,a_n)$ then $G(a)\leq A(a)$.

Proof: Let us note first that the theorem holds for $n=2$. Indeed, $$(a_1-a_2)^2=a_1^2-2a_1a_2+a_2^2\geq0,$$ so $$(a_1+a_2)^2\geq4a_1a_2,$$ with equality iff $a_1=a_2$.

Suppose now that the theorem holds for $n=m$. We shall show that the theorem holds for $n=2m$. Let $a_1,\ldots a_m,b_1,\ldots,b_m$ be non-negative reals. Then: $$(a_1\cdots a_mb_1\cdots b_m)^{1/2m}=\left((a_1\cdots a_m)^{1/m}(b_1\cdots b_m)^{1/m}\right)^{1/2}$$ $$\leq\frac{1}{2}\left(a_1\cdots a_m)^{1/m}+(b_1\cdots b_m)^{1/m}\right)$$ $$\leq\frac{1}{2}\left(\frac{a_1+\cdots+a_m}{m}+\frac{b_1+\cdots b_m}{m}\right)=\frac{a_1+\cdots+a_m+b_1+\cdots b_m}{2m}$$ If equality holds then, by the induction hypothesis, we have $a_1=\cdots=a_m=b_1=\cdots=b_m$. This implies that the theorem holds whenever $n$ is a power of $2$.

Finally, suppose $n$ is an arbitrary integer. Let $n<2^k=N$ and $a=\frac{1}{n}\sum_{i=1}^n a_i$. Set $a_{n+1}=\cdots=a_N=a$. Then: $$\prod_{i=1}^N a_i=a^{N-n}\prod_{i=1}^n a_i\leq\left(\frac{1}{N}\sum_{i=1}^N a_i\right)^N=a^N,$$ so: $$\prod_{i=1}^n a_i\leq a^n,$$ with equality iff $a_1=\cdots=a_N$, in other words iff $a_1=\cdots=a_n$. $\square$

Answer 2

Note that geometric means states

The geometric mean applies only to positive numbers.

However, if you allow negative values, as long as the product is positive, then there are solutions. To see this, let $a$, $b$ and $c$ be positive integers. Next, have the $8$ integers involved be $-a$, $a$, $-b$, $b$, $2c$, $4c$, $6c$ and $12c$. Then the arithmetic mean would be

$$\begin{equation}\begin{aligned} a & = \frac{-a + a - b + b + 2c + 4c + 6c + 12c}{8} \\ & = \frac{24c}{8} \\ & = 3c \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

and the geometric mean would be

$$\begin{equation}\begin{aligned} g & = \sqrt[8]{(-a)(a)(-b)(b)(2c)(4c)(6c)(12c)} \\ & = \sqrt[8]{a^2b^2c^4(2^6)(3^2)} \\ & = \sqrt[4]{abc^2(2^3)(3)} \end{aligned}\end{equation}\tag{2}\label{eq2A}$$

Equating these $2$ values and taking each side to the $4$th power gives

$$\begin{equation}\begin{aligned} (3^4)c^4 & = abc^2(2^3)(3) \\ (3^3)c^2 & = ab(2^3) \end{aligned}\end{equation}\tag{3}\label{eq3A}$$

Note that one solution, among many, to this is $a = 3^2 = 9$, $b = 2(3) = 6$, $c = 2^2 = 4$. Thus, a set of $8$ integers would then be $\{-9,9,-6,6,8,16,24,48\}$.

I don't know if there are any solutions involving fewer than $8$ integers, but this is the first one I found. Also, there are many additional cases involving $8$ or more integers as well, including values for \eqref{eq3A}.