Find the smallest $n$ for which there are real $a_{1}, a_{2}, \ldots,a_{n}$ such that

$$\left\{\begin{array}{l} a_{1}+a_{2}+\ldots+a_{n}>0 \\a_{1}^{3}+a_{2}^{3}+\ldots+a_{n}^{3}<0 \\a_{1}^{5}+a_{2}^{5}+\ldots+a_{n}^{5}>0 \end{array}\right.$$

I don't have a solution, but I suspect $n_{\min}=5$ is optimal. I think so mainly because I could not find anything by brute force for either $4$ or $3$. There are quite a few solutions for $n=5$, for example:

$$a_{1}=1, a_{2}=-0.88, a_{3}=-0.88, a_{4}=0.5, a_{5}=0.5$$

$$a_{1}=-8, a_{2}=-7, a_{3}=3, a_{4}=4, a_{5}=9$$

It is also not difficult to prove by contradiction that $n_{\min}>2$.


2 Answers 2


Claim. $n_\min=5$.

Proof: At least one element must be positive, so the system can be rewritten as $$\left\{\begin{array}{l} 1+b_{1}+\cdots+b_{n-1}>0 \\1+b_{1}^{3}+\cdots+b_{n-1}^{3}<0 \\1+b_{1}^{5}+\cdots+b_{n-1}^{5}>0 \end{array}\right.$$ where $b_i=a_{i+1}/a_1$. Intuitively, there should be an infinite number of solutions when $n>4$ as there are more variables than inequalities.

When $n=1$ there is nothing to prove.

When $n=2$, the inequalities $1+b_1>0$ and $1+b_1^3<0$ contradict each other.

When $n=3$, we have $-(1+b_1^5)^{1/5}<b_2<-(1+b_1^3)^{1/3}$ and $-(1+b_1)<b_2<-(1+b_1^3)^{1/3}$. The first double inequality is equivalent to $(1+b_1^5)^3>(1+b_1^3)^5$ which is true only when $b_1\in(-1,0)$; this can be shown by considering the sign of $$f'(b_1)=15b_1^2(b_1+1)^2(2b_1^2-2b_1+1)(2b_1^5-2b_1^4+2b_1^3+1)$$ where $f(b_1)=(1+b_1^3)^5-(1+b_1^5)^3$. The second double inequality is equivalent to $(1+b_1)^3>1+b_1^3$ which is false when $b_1\in(-1,0)$; contradiction.

When $n=4$, we have \begin{align}-(1+b_1^5+b_2^5)^{1/5}<b_3&<-(1+b_1^3+b_2^3)^{1/3}\\-(1+b_1+b_2)<b_3&<-(1+b_1^3+b_2^3)^{1/3}.\end{align} The second double inequality is equivalent to $(1+b_1+b_2)^3>1+b_1^3+b_2^3$, or that $$(b_1+b_2)(b_1+1)(b_2+1)>0\implies\left(c_1^{1/5}+c_2^{1/5}\right)\left(c_1^{1/5}+1\right)\left(c_2^{1/5}+1\right)>0$$ where $c_1=b_1^5$ and $c_2=b_2^5$.

The first double inequality is equivalent to $g(c_1,c_2)=\left(1+c_1^{3/5}+c_2^{3/5}\right)^{5/3}-(1+c_1+c_2)<0$. When $c_2>-1$, we have the criterion $\left(c_1^{1/5}+c_2^{1/5}\right)\left(c_1^{1/5}+1\right)>0$. Now the partial derivative $$g_{c_1}(c_1,c_2)=c_1^{-2/5}\left(1+c_1^{3/5}+c_2^{3/5}\right)^{2/3}-1$$ gives the stationary point $c_1^*=-2^{-5/3}\left(1+c_2^{3/5}\right)^{5/3}$ which we can show is the global minimum by considering second derivatives. In addition, we have $$g(-c_2,c_2)=g(-1,c_2)=0,$$ and $-c_2,-1$ are the only roots since $g_{c_1}$ is strictly decreasing on $(-\infty,c_1^*)$ and increasing on $(c_1^*,\infty)$. Hence $g(c_1,c_2)<0$ only when $c_1\in(\min\{-1,-c_2\},\max\{-1,-c_2\})$, which contradicts the criterion.

When $c_2<-1$, we have the criterion $\left(c_1^{1/5}+c_2^{1/5}\right)\left(c_1^{1/5}+1\right)<0$. But then, in a similar vein to above, $g(c_1,c_2)<0$ only when $c_1\in\Bbb R\setminus(\min\{-1,-c_2\},\max\{-1,-c_2\})$ and we immediately obtain the contradiction. $\square$

Remark 1. When $n=5$, suppose that $b_1=b_3$ and $b_2=b_4$ solve the system. Akin to the proof of the case $n=3$, we have the two inequalities $(1/2+b_1^5)^3>(1/2+b_1^3)^5$ and $(1/2+b_1)^3>1/2+b_1^3$. Numerically, both inequalities hold in the interval $$b_1\in(-0.897,-0.809)\cup(0.309,0.605)$$ which explains the first set of your solutions.

Remark 2. The proof in the case $n=4$ can be greatly reduced in length if we can prove the following generalisation:

If $p\in(0,1)$ is a rational number with an odd numerator and denominator, then the inequality $1+x^p+y^p<(1+x+y)^p$ has solution $(1+x)(1+y)(x+y)<0$ for all such $p$.

This should probably be investigated in a separate question. For the purposes of this problem, the inequality $g(c_1,c_2)<0$ is equivalent to the above with $p=3/5$, where the inequality $(1+c_1)(1+c_2)(c_1+c_2)<0$ would contradict $\left(c_1^{1/5}+c_2^{1/5}\right)\left(c_1^{1/5}+1\right)\left(c_2^{1/5}+1\right)>0$.

  • $\begingroup$ To check, how does finding the signage of $f'(b)$ tell you about when $f(b) > 0$? (I showed that the real roots of $f(b) = $-1, 0$, from which the result follows.) $\endgroup$
    – Calvin Lin
    Dec 29, 2021 at 23:52
  • $\begingroup$ The formulation of the generalized problem in the 2nd remark is not clear to me. $\endgroup$
    – QLimbo
    Dec 30, 2021 at 0:09
  • $\begingroup$ @CalvinLin You are correct that $f(-1)=f(0)=0$. To show that these are the only roots, I considered the derivative. As described above, the sign of the derivative depends only on the sign of $2b^5-2b^4+2b^3+1$, which is strictly increasing (except $b=0$ where it is an inflection point) with one root at $\approx-0.6$, and the conclusion follows. $\endgroup$
    – TheSimpliFire
    Dec 30, 2021 at 9:56
  • $\begingroup$ @Cornifer See the last paragraph of revised. $\endgroup$
    – TheSimpliFire
    Dec 30, 2021 at 10:01

n = 3 seems to be fine. Let $c$ be a parameter to be determined later. Put $a_1 =-1-6c^3$, $a_2=-1+6c^3$, and $a_3=6c^2$. Clearly $a_1 + a_2 + a_3 = -1 +6c^2$ which is positive for $c$ large enough. Now, $a_1^3 + a_2^3 +a_3^3 = (-1-6c^3)^3 + (-1 + 6c^3)^3 + (6c^2)^3 = -2 < 0 $ Also, $a_1^5 + a_2^5 + a_3^3 = (-1 - 6c^3)^5 + (-1 + 6c^3)^5 + (6c^2)^5$ has the leading coefficient $2*6^4*c^{12}$ that is it, it is positive for $c$ large enough.

  • $\begingroup$ Nice try but it doesn't work. Check the leading coefficient in the expansion of the sum of the $5$-th powers. $\endgroup$
    – quasi
    Dec 29, 2021 at 20:03
  • $\begingroup$ Everything is fine - 5th powers cancel out. $\endgroup$
    – Salcio
    Dec 29, 2021 at 21:07
  • $\begingroup$ in the sum of 5th powers, the leading term is -12960 $c^{12}$. $\endgroup$
    – cineel
    Dec 29, 2021 at 21:13
  • $\begingroup$ Of course it is ... $\endgroup$
    – Salcio
    Dec 29, 2021 at 21:18

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