For $n \geq 2$, prove or disprove that $1 < \frac {x_1 + x_2 +... + x_n}{n} \leq 2$ ,

for all natural numbers $x_1,x_2,...,x_n$ statisfying

$${\sum \limits_{j=1}^{n}{x_j}} = {\prod \limits_{j=1}^{n}{x_j}}.$$

EDIT : As per the request of @yurnero, I hereby explicitly state that the "natural numbers" mentioned in the question mean the positive non-fraction numbers excluding $0$.

My attempt :

Let ${\sum \limits_{j=1}^{n}{x_j}}={\prod \limits_{j=1}^{n}{x_j}}=\lambda$.

Applying AM-GM on the set {$x_1,x_2,...,x_n$} :

$$\frac {\lambda}{n} \geq {\lambda}^{\frac {1}{n}}$$

$$\frac {{\lambda}^n}{n^n} \geq {\lambda}$$

$${\lambda ^{n-1}} \geq n^{n-1} . n$$

$$\lambda \geq n . n^{\frac {1}{n-1}}$$

$$\frac {\lambda}{n} \geq n^{\frac {1}{n-1}}$$

Is this approach correct? If yes, then is there any alternate way of "solving" this problem ? How to solve for the other bound/limit ?

  • $\begingroup$ Your approach is not correct. I think you mean $\lambda/2 \geq 2$. Even in this case ($n=2$), $\lambda/2$ can be equal to $2$, which does not disprove the inequality. $\endgroup$
    – Alex Silva
    Jan 18, 2017 at 13:03
  • $\begingroup$ @AlexSilva Oh yeah.. i'll edit accordingly... anyways how to prove the inequality,,,???? $\endgroup$
    – user399078
    Jan 18, 2017 at 13:06
  • 1
    $\begingroup$ Are the $x_i$ non-negative integers? Then $\frac {x_1 + x_2 +... + x_n}{n} > 1$ would be (almost) trivial. $\endgroup$
    – Martin R
    Jan 18, 2017 at 13:23
  • 1
    $\begingroup$ The lower inequality isn't true when $x_1=\ldots=x_n=0$ so you might want to clarify that your definition of natural numbers excludes $0$. $\endgroup$
    – yurnero
    Jan 18, 2017 at 13:44
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    $\begingroup$ There are 2 conventions. See: en.wikipedia.org/wiki/Natural_number. In particular: "Some definitions, including the standard ISO 80000-2, begin the natural numbers with 0 ..." You should clarify which you use. No need for shouting, please. $\endgroup$
    – yurnero
    Jan 18, 2017 at 13:52

1 Answer 1


If $x_1+x_2+\cdot+x_n=x_1x_2\cdots x_n$ with $n\ge2$, it's easy to see that at least two of the $x_i$'s must be greater than $1$. Suppose we reorder things so that $x_i=2+u_i$ with $u_i\ge0$ for $1\le i\le k$ and $x_i=1$ for $k+1\le i\le n$. If we now let

$$P=(2+u_1)(2+u_2)\cdots(2+u_k)\quad\text{and}\quad S=(2+u_1)+(2+u_2)+\cdots+(2+u_k)$$

then we have $P=S+1+1+\cdots+1$ with $n-k$ $1$'s, from which we see that $n=k+P-S$, and the inequalities to prove become

$$1\le{P\over k+P-S}\le2$$


$$k\le S\quad\text{and}\quad 2(S-k)\le P$$

The first of these is clearly true, since $S=(2+u_1)+(2+u_2)+\cdots+(2+u_k)\ge2+2+\cdots+2=2k$. The second requires our initial observation that $k\ge2$:

$$\begin{align} 2(S-k) &=2k+2(u_1+u_2+\cdots+u_k)\\ &\le2^k+2^{k-1}(u_1+u_2+\cdots+u_k)\\ &\le(2+u_1)(2+u_2)\cdots(2+u_k)\\ &=P \end{align}$$


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