Question 60
Suppose $a, b, c$ are real numbers such that $a+b \geq 0$, $b+c \geq 0$, and $c+a \geq 0$. Prove that
$$
a+b+c \geq \frac{|a|+|b|+|c|}{3} .
$$
(Note: $|x|$ is called the absolute value of $x$ and is defined as follows. If $x \geq 0$ then $|x|=x$; and if $x\lt 0$ then $|x|=-x$. For example, $|6|=6$, $|0|=0$ and $|-6|=6$.)
$$
a+b+c \geq \frac{|a|+|b|+|c|}{3} .
$$
(Note: $|x|$ is called the absolute value of $x$ and is defined as follows. If $x \geq 0$ then $|x|=x$; and if $x\lt 0$ then $|x|=-x$. For example, $|6|=6$, $|0|=0$ and $|-6|=6$.)