IMO 2015 Team Selection Tests

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

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 $$a_1+a_2+a_3>b_4,a_1^3+a_2^3+a_3^3<b_4^3,a_1^5+a_2^5+a_3^5>b_4^5.$$ 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 $$x,y,z,t>0,x+y>z+t,x^3+y^3<z^3+t^3,x^5+y^5>z^5+t^5.$$ W.l.o.g, we assume that $x\ge y$ and $z\ge t$. Since $$x^3+y^3=(x+y)(x^2-xy+y^2)<(z+t)(z^2-zt+t^2)=z^3+t^3,$$ and $x+y>z+t$, so $z^2-zt+t^2>x^2-xy+y^2$. 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 $$z-x>\frac{y^3-t^3}{z^2+zx+x^2}(1).$$ From $x^5+y^5>z^5+t^5$, we have $y^5-t^5>z^5-x^5$. Together with (1), we have $$y^5-t^5>(y^3-t^3)\frac{z^4+z^{3}x+z^2{x}^2+z{x}^3+x^4}{z^2+zx+x^2}.$$ Hence, $$(y^4+y^{3}t+y^2{t}^2+y{t}^3+t^4)(z^2+zx+x^2)>(y^2+yt+t^2)(z^4+z^{3}x+z^2{x}^2+z{x}^3+x^4).$$ Set $X=z^2+zx+x^2$ and $Y=y^2+yt+t^2$, 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 $$xyzt(xz-yt)>(x^2-y^2)(XY-z^2{t}^2)+(z^2-t^2)(XY-x^2{y}^2).$$ The last inequality is a contradiction as $x^2-y^2\ge 0$, $z^2-t^2>zx-yt$ and $$XY-x^2{y}^2=(x^2+xz+z^2)(y^2+yt+t^2)-x^2{y}^2>xyzt.$$ Therefore, the minimum positive integer $n$ satisfying the given conditions is $n=5$.