XVI Junior Macedonian Mathematical Olympiad

At first we notice that the equality $$\begin{array}{rl}(a+1)(b+1)+(a+1)(c+1)+(b+1)(c+1)& =a+b+c+(a+b+c+2)+ab+ac+bc+1\\ & =a+b+c+abc+ab+ac+bc+1=(a+1)(b+1)(c+1)\end{array}$$ holds. Now from the inequality between the arithmetic and geometric mean we get: $$\begin{array}{rl}\frac{a}{b+1}+\frac{b}{c+1}+\frac{c}{a+1}& =\frac{a+1}{b+1}+\frac{b+1}{c+1}+\frac{c+1}{a+1}-\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)\\ & \ge 3\\ & \ge 3\sqrt{\frac{(a+1)(b+1)(c+1)}{(b+1)(c+1)(a+1)}}-\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)=3\\ & =3-\frac{(a+1)(b+1)(c+1)}{(a+1)(b+1)(c+1)}=3-1=2\end{array}$$ Equality holds if and only if $a=b=c=2$.

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Alternative solution.

If we multiply the inequality by $(a+1)(b+1)(c+1)$ we get the equivalent inequality: $$a(a+1)(c+1)+b(b+1)(a+1)+c(c+1)(b+1)\ge 2(a+1)(b+1)(c+1)$$ which can be written in the form $$a^{2}c+b^{2}a+c^{2}b+a^2+b^2+c^2\ge 2abc+ab+bc+ca+a+b+c+2.$$ But, from the inequalities $$a^{2}c+b^{2}a+c^{2}b\ge 3\sqrt[3]{a^3{b}^3{c}^3}=3abc$$ $$a^2+b^2+c^2\ge ab+bc+ca$$ we get $$a^{2}c+b^{2}a+c^{2}b+a^2+b^2+c^2\ge 3abc+ab+bc+ca.$$ Now, if we use the condition of the exercise, we get $$a^{2}c+b^{2}a+c^{2}b+a^2+b^2+c^2\ge 2abc+ab+bc+ca+a+b+c+2.$$