SELECTION EXAMINATION

The inequality is equivalent to $$\frac{c(a+1)+a(b+1)+b(c+1)}{abc(a+1)(b+1)(c+1)}\ge \frac{3}{(1+abc{)}^2},$$ or after simplifications to $$(1+abc{)}^2(ab+bc+ca+a+b+c)\ge 3abc(ab+bc+ca+a+b+c+abc+1).$$ We put $m=a+b+c$, $n=ab+bc+ca$ and $x^3=abc$, to obtain: $$(m+n)(1+x^3{)}^2\ge 3x^3(x^3+m+n+1),$$ or $(m+n)(x^6-x^3+1)\ge 3x^3(x^3+1).$ From the inequality of arithmetic–geometric mean we have $m\ge 3x$ and $n\ge 3x^2$, and so $m+n\ge 3x(x+1)$. Therefore, it is enough to prove that $$\begin{array}{rl}& 3x(x+1)(x^6-x^3+1)\ge 3x^3(x+1)(x^2-x+1)\\ \iff & x^6-x^3+1\ge x^2(x^2-x+1)\\ \iff & x^6-x^4-x^2+1\ge 0\iff x^4(x^2-1)-(x^2-1)\ge 0\\ \iff & (x^2-1)(x^4-1)\ge 0\iff (x^2+1)(x^2-1{)}^2\ge 0,\end{array}$$ which is valid.