Macedonian Junior Mathematical Olympiad

By twice using the inequality between the arithmetical mean and geometrical mean we get $$\begin{array}{rl}(16a^2+8b+17)& =(16a^2+1+8b+16)\ge 8a+8b+16=8(a+b+2)=8(a+1+b+1)\ge \\ & \ge 8\cdot 2\sqrt{(a+1)(b+1)}=2^4\sqrt{(a+1)(b+1)}.\end{array}(1)$$ Analogously we have $$(16b^2+8c+17)\ge 2^4\sqrt{(b+1)(c+1)}(2)$$ $$(16c^2+8a+17)\ge 2^4\sqrt{(c+1)(a+1)}(3).$$ If we multiply the three inequalities we get $$(16a^2+8b+17)(16b^2+8c+17)(16c^2+8a+17)\ge 2^{12}(a+1)(b+1)(c+1).$$ In (1) equality is obtained when $16a^2=1$ and $a=b$, i.e. $a=b=\frac{1}{4}$. By an analogous argument for (2) and (3) we get $a=b=c=\frac{1}{4}$.