Balkan Mathematical Olympiad Shortlist

The proof is essentially a size argument. We split into three cases, the first two of which are quite straightforward.

Case 1: two of $p,q,r$ are equal, wlog $q=r$. Subtract $a^q+b^q+c^q$ from the given equation to show that $$a^p-a^q=b^p-b^q=c^p-c^q.$$ Now if $p>q$ then $x^p-x^q=x^q(x^{p-q}-1)$ is a strictly increasing function of $x>0$, so the only way the equality can hold is if $a=b=c$. If $p<q$ then the same argument shows that $a=b=c$ also. Yet the only remaining subcase is when $p=q=r$.

Case 2: two of $a,b,c$ are equal, wlog $b=c$. Subtract $b^p+b^q+b^r$ from the given equation to show that $a^p-b^p=a^q-b^q=a^r-b^r$. Exactly as in the previous case, the function of a positive integer $s$ $a^s-b^s=(a-b)(a^{s-1}+a^{s-2}b+\cdots +b^{s-1})$ is strictly increasing, strictly decreasing or zero according as $a>b$, $a<b$ or $a=b$. In the first two subcases, this forces $p=q=r$, and in the last subcase $a=b=c$.

Case 3: the $a,b,c$ are distinct, as are the $p,q,r$. Wlog $a$ is the greatest of $a,b,c$ and (cycling the variables if necessary) $p$ is the greatest of $p,q,r$. In particular, $a,p\ge 3$. We claim that for such $a,p$ we have $$a^p\ge (a-1{)}^p+2a^{p-1}.$$ Indeed, since $(a-1{)}^p+2a^{p-1}\le a^{p-3}((a-1{)}^3+2a^2)$ it suffices to prove the inequality for $p=3$, when it rearranges to the inequality $a^2-3a+1\ge 0$. This certainly holds for $a\ge 3$. As a consequence, we have the inequality $$a^p+b^q+c^r>(a-1{)}^p+2a^{p-1}\ge b^p+c^q+a^r$$ which is a contradiction.