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

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 $$n_{\min}>2$$.

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 $$-(1+b_1^5)^{1/5} and $$-(1+b_1). 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} 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 $$\left(c_1^{1/5}+c_2^{1/5}\right)\left(c_1^{1/5}+1\right)\left(c_2^{1/5}+1\right)>0$$.

• 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$, from which the result follows.) Dec 29, 2021 at 23:52 • The formulation of the generalized problem in the 2nd remark is not clear to me. Dec 30, 2021 at 0:09 • @CalvinLin You are correct that$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. Dec 30, 2021 at 9:56 • @Cornifer See the last paragraph of revised. 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. • Nice try but it doesn't work. Check the leading coefficient in the expansion of the sum of the$5$-th powers. Dec 29, 2021 at 20:03 • Everything is fine - 5th powers cancel out. Dec 29, 2021 at 21:07 • in the sum of 5th powers, the leading term is -12960$c^{12}\$. Dec 29, 2021 at 21:13
• Of course it is ... Dec 29, 2021 at 21:18