, 2015

For an $n$-tuple $\mathbf{a} = (a_1, a_2, \dots, a_n)$ and a nonnegative integer $k$, let
$$S_k(\mathbf{a}) = a_1^k + \dots + a_n^k.$$

We need to find the smallest positive integer $n$ such that there exist an $n$-tuple $\mathbf{a} = (a_1, \dots, a_n)$ of real numbers such that $S_1(\mathbf{a}) > 0$, $S_3(\mathbf{a}) < 0$ and $S_5(\mathbf{a}) > 0$.
Note that if there exists an $n$-tuple satisfying the given condition then for any $m > n$, by adding extra $m-n$ zeros, there exists a $m$-tuple satisfying the given condition.
First we show that there exists a 5-tuple $\mathbf{a} = (a_1, a_2, a_3, a_4, a_5)$ such that $S_1(\mathbf{a}) > 0$, $S_3(\mathbf{a}) < 0$, and $S_5(\mathbf{a}) > 0$. We choose $a_1 = 2x, a_2 = a_3 = 1, a_4 + a_5 = -(x+1)$ then $S_1(\mathbf{a}) = 0$. We choose $x$ so that $S_3(\mathbf{a}) = 8x^3 + 2 - 2(x+1)^3 < 0$ and $S_5(\mathbf{a}) = 32x^5 + 2 - 2(x+1)^5 > 0$. Solving these inequalities, we can take $x = 1.5$. After that, we adjust $a_4, a_5$ to obtain a 5-tuple satisfying the conditions: $S_1(\mathbf{a}) > 0$, $S_3(\mathbf{a}) < 0$, and $S_5(\mathbf{a}) > 0$ as follows $a_1 = 3, a_2 = a_3 = 1, a_4 = a_5 = -2.45$.

Now we show that there does not exist a 4-tuple $\mathbf{a} = (a_1, a_2, a_3, a_4)$ such that $S_1(\mathbf{a}) > 0$, $S_3(\mathbf{a}) < 0$, and $S_5(\mathbf{a}) > 0$. Suppose for the contrary, there exists a 4-tuple $\mathbf{a} = (a_1, a_2, a_3, a_4)$ such that $S_1(\mathbf{a}) > 0$, $S_3(\mathbf{a}) < 0$, and $S_5(\mathbf{a}) > 0$, then there is at least one positive and one negative among the $a_i$'s. We consider three cases.

Case 1. There is one positive and three non-positive among $a_i$'s. Suppose that $a_1 > 0 \ge a_2, a_3, a_4$. Let $b_i = -a_i$ for $i = 2, 3, 4$ then $a_1 > b_2 + b_3 + b_4$. This implies that
$$a_1^3 > (b_2 + b_3 + b_4)^3 \ge b_2^3 + b_3^3 + b_4^3 = -(a_2^3 + a_3^3 + a_4^3),$$
which is a contradiction.

Case 2. There are three non-negative and one negative among $a_i$'s. Suppose that $a_1, a_2, a_3 \ge 0 > a_4$. Let $b_4 = -a_4$ then
$$a_1 + a_2 + a_3 > b_4, \quad a_1^3 + a_2^3 + a_3^3 < b_4^3, \quad a_1^5 + a_2^5 + a_3^5 > b_4^5.$$
From the second inequality, we have $a_1, a_2, a_3 < b_4$. Hence,
$$a_1^5 + a_2^5 + a_3^5 < a_1^3 b_4^2 + a_2^3 b_4^2 + a_3^3 b_4^2 = (a_1^3 + a_2^3 + a_3^3) b_4^2 < b_4^5,$$
which is a contradiction.

Case 3. There are two positive and two negative among $a_i$'s. Let two positive numbers be $x, y$ and two negative numbers be $-z, -t$ then we have
$$x, y, z, t > 0, \quad x + y > z + t, \quad x^3 + y^3 < z^3 + t^3, \quad x^5 + y^5 > z^5 + t^5.$$
W.l.o.g, we assume that $x \ge y$ and $z \ge t$. Since
$$x^3 + y^3 = (x + y)(x^2 - x y + y^2) < (z + t)(z^2 - z t + t^2) = z^3 + t^3,$$
and $x + y > z + t$, so $z^2 - z t + t^2 > x^2 - x y + y^2$. This implies that
$$(z + t)^2 + 3(z - t)^2 > (x + y)^2 + 3(x - y)^2.$$
Therefore, $z - t > x - y$. If $z \le x$, then $y - t > x - z \ge 0$ or $y > t$. Hence, $x^3 + y^3 > z^3 + t^3$, which is a contradiction. So, we have $z > x \ge y > t$.
From $x^3 + y^3 < z^3 + t^3$, we have $z^3 - x^3 > y^3 - t^3$, or
$$z - x > \frac{y^3 - t^3}{z^2 + z x + x^2} \quad (1).$$
From $x^5 + y^5 > z^5 + t^5$, we have $y^5 - t^5 > z^5 - x^5$. Together with (1), we have
$$y^5 - t^5 > (y^3 - t^3) \frac{z^4 + z^3 x + z^2 x^2 + z x^3 + x^4}{z^2 + z x + x^2}.$$
Hence,
$$(y^4 + y^3 t + y^2 t^2 + y t^3 + t^4)(z^2 + z x + x^2) > (y^2 + y t + t^2)(z^4 + z^3 x + z^2 x^2 + z x^3 + x^4).$$
Set $X = z^2 + z x + x^2$ and $Y = y^2 + y t + t^2$, then the above inequality can be rewritten as
$$((y^2 + t^2)Y - y^2 t^2)X > ((z^2 + x^2)X^2 - z^2 x^2)Y,$$
or
$$xyzt(xz - y t) > (x^2 - y^2)(X Y - z^2 t^2) + (z^2 - t^2)(X Y - x^2 y^2).$$
The last inequality is a contradiction as $x^2 - y^2 \ge 0$, $z^2 - t^2 > z x - y t$ and
$$X Y - x^2 y^2 = (x^2 + x z + z^2)(y^2 + y t + t^2) - x^2 y^2 > x y z t.$$
Therefore, the minimum positive integer $n$ satisfying the given conditions is $n = 5$.