Mathematical Reflections 2017, Issue 1 - Problem J400

Problem:
Prove that for all real numbers $a,b,c$ the following inequality holds:
$$\dfrac{|a|}{1+|b|+|c|}+\dfrac{|b|}{1+|c|+|a|}+\dfrac{|c|}{1+|a|+|b|} \geq \dfrac{|a+b+c|}{1+|a+b+c|}.$$
When does the equality occur?

Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam

Solution:
 Put $s=a+b+c$. By Triangle Inequality, we have $|s| \leq |a|+|b|+|c|$, so $|s|(1+|a|+|b|+|c|) \leq (1+|s|)(|a|+|b|+|c|)$, i.e.
$$\dfrac{|s|}{1+|s|} \leq \dfrac{|a|+|b|+|c|}{1+|a|+|b|+|c|}.$$ Using the fact that $|x| \geq 0$ for all real numbers $x$, we have
$$\begin{array}{lll} \dfrac{|a|+|b|+|c|}{1+|a|+|b|+|c|}&=&\dfrac{|a|}{1+|a|+|b|+|c|}+\dfrac{|b|}{1+|a|+|b|+|c|}+\dfrac{|c|}{1+|a|+|b|+|c|} \\ & \leq & \dfrac{|a|}{1+|b|+|c|}+\dfrac{|b|}{1+|c|+|a|}+\dfrac{|c|}{1+|a|+|b|}. \end{array}$$
So, $$\dfrac{|s|}{1+|s|} \leq \dfrac{|a|}{1+|b|+|c|}+\dfrac{|b|}{1+|c|+|a|}+\dfrac{|c|}{1+|a|+|b|},$$ which is the desired inequality. Equality occurs if and only if $|a|=|b|=|c|=0$, i.e. if and only if $a=b=c=0$.