SAUDI ARABIAN MATHEMATICAL COMPETITIONS

First, we will show that $a$, $b$, $c$ are pairwise coprime. Denote $d=\gcd (b,c)$ and suppose that $d>1$. Take $p$ as a prime divisor of $d$. We have $$\{\begin{array}{l}p\mid b\\ p\mid c\end{array}\Rightarrow \{\begin{array}{l}p\mid (c-a{)}^2\\ p\mid (a-b{)}^2\end{array}\Rightarrow \begin{array}{l}p\mid c-a\\ p\mid a-b\end{array}\Rightarrow p\mid a.$$ This implies that $p\mid \gcd (a,b,c)$, a contradiction. So $\gcd (b,c)=1$. Similar with $\gcd (c,a)$, $\gcd (a,b)$.

Suppose that $a$, $b$, $c$ are side-lengths of some triangle then $$b+c>a,c+a>b,a+b>c.$$ So by putting $x=(a+b-c)(b+c-a)(c+a-b)>0$, we can see that $x>0$.

We have $a\mid (b-c{)}^2-a^2=(b-c-a)(b-c+a)$, thus $a\mid x$. Similarly, we have $b\mid x$, $c\mid x$ and since $\gcd (a,b)=\gcd (b,c)=\gcd (c,a)=1$ so $abc\mid x$. Hence, $$abc\le (a+b-c)(b+c-a)(c+a-b).$$ From AM-GM, we have $(a+b-c)(b+c-a)\le {\left(\frac{a+b-c+b+c-a}{2}\right)}^2=b^2$ and similar inequalities as $(b+c-a)(c+a-b)\le c^2$, $(c+a-b)(a+b-c)\le a^2$ then $$(a+b-c)(b+c-a)(c+a-b)\le abc.$$ These imply the equality must be hold, so $a=b=c$, contradiction since $a$, $b$, $c$ pairwise distinct. $\square$