Mongolian Mathematical Olympiad

It is possible to prove that for odd natural number $n>1$ and for arbitrary real numbers $a$, $b$, $c$ which satisfy the condition $a+b+c=0$, the numbers $abc$ and $a^n+b^n+c^n$ have the same sign. It is obvious that after setting $n=5$, $a=x-y$, $b=y-z$, $c=z-x$ follows required statement. Let us use following property. Property. a) If $x$, $y>0$ and $k>1$ then $x^k+y^k<(x+y{)}^k$; b) If $x$, $y<0$ and $k>1$, $k$ odd then $x^k+y^k>(x+y{)}^k$.

First consider the case that one of $a$, $b$, $c$ is equal to $0$. For example, if $a=0$ then $abc=0$ and $a+b+c=b+c=0$. From this we get $b=-c$ and $a^n+b^n+c^n=b^n+c^n=(-c{)}^n+c^n=0$. In this case nothing to prove. Now consider the case $abc\ne 0$. Without losing generality it is possible to assume $a\ge b\ge c$. From $a+b+c=0$ follows $a>0$ and $c<0$. i) $b<0\Rightarrow abc<0$ and $a^n+b^n+c^n<(a+b{)}^n+c^n=(-c{)}^n+c^n=0$ by property a). ii) $b<0\Rightarrow abc>0$ and $a^n+b^n+c^n>a^n+(b+c{)}^n=a^n+(-a{)}^n=0$ by property b).