Fall Mathematical Competition

Taking $a=b=c$ we obtain $$r{a}^2+(3-r)\frac{1}{a}\ge 3⟺(a-1)(r(a^2+a+1)-3)\ge 0$$ for any $a>0$. Then it easily follows that $r=1$.

Conversely, let $r=1$. Then we write the inequality as $\frac{2+abc}{3}\ge \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$. The GM-HM inequality implies that the right hand side does not exceed $\sqrt[3]{abc}$ and, setting $x=\sqrt[3]{abc}>0$, it is enough to prove that $$\frac{2+x^3}{3}\ge x$$ which is equivalent to the obvious $(x-1{)}^2(x+2)\ge 0$.