Romanian Mathematical Olympiad

The inequality is equivalent to $$a^2+b^2+c^2-a-b-c\le (a+b+c{)}^2-9(a+b+c)+24,$$ or $$0\le ab+bc+ca-4(a+b+c)+12.$$ Since $ab-2a-2b+4=(a-2)(b-2)$, summing together with the similar relations implies $(ab-2a-2b+4)+(bc-2b-2c+4)+(ca-2c-2a+4)=(a-2)(b-2)+(b-2)(c-2)+(c-2)(a-2)\ge 0$, since $a$, $b$, $c\ge 2$.