Macedonian Mathematical Olympiad

From the inequality between the arithmetic and geometric means we have $$\frac{a^3+2}{b+2}=\frac{a^3+1+1}{b+2}\ge \frac{3\sqrt[3]{a^3\cdot 1\cdot 1}}{b+2}=\frac{3a}{b+2}.$$ Analogously we get the equations $$\frac{b^3+2}{c+2}\ge \frac{3b}{c+2}\text{and}\frac{c^3+2}{a+2}\ge \frac{3c}{a+2}.$$ Now we have $$\frac{a^3+2}{b+2}+\frac{b^3+2}{c+2}+\frac{c^3+2}{a+2}\ge 3\left(\frac{a}{b+2}+\frac{b}{c+2}+\frac{c}{a+2}\right)(1)$$ Then, according to the Cauchy-Bunyakovsky inequality we have $$\begin{array}{rl}\frac{a}{b+2}+\frac{b}{c+2}+\frac{c}{a+2}& =\frac{a^2}{a(b+2)}+\frac{b^2}{b(c+2)}+\frac{c^2}{c(a+2)}\ge \\ & \ge \frac{(a+b+c{)}^2}{a(b+2)+b(c+2)+c(a+2)}=\frac{(a+b+c{)}^2}{ab+bc+ca+2(a+b+c)}\end{array}(2)$$ From the obvious inequality $(a-b{)}^2+(b-c{)}^2+(c-a{)}^2\ge 0$ follows the inequality $$(a+b+c{)}^2\ge 3(ab+bc+ca)\text{i.e.}\frac{1}{ab+bc+ca}\ge \frac{3}{(a+b+c{)}^2}(3)$$

Now from (2) and (3) we get Finally, from (1), (4) and the condition $a+b+c=3$ we get which completes the proof. Equality holds if and only if $a=b=c=1$.