Solution:
The inequality $b+c \geq 0$ gives $a+b+c \geq a$. On the other hand, adding up the other two given inequalities yields $(a+b)+(c+a) \geq 0$, resulting in $a+b+c \geq -a$. Since $|a|=a$ or $-a$, we have in any case that
$$a+b+c \geq |a| .$$
Similarly
$$\begin{aligned} & a+b+c \geq |b| \\ & a+b+c \geq |c| \end{aligned}$$
Now adding these three inequalities and dividing by 3 yields the desired inequality.

Alt Solution 1:
The previous solution used the symmetry of $a, b$, and $c$. We can also use that symmetry to assume without loss of generality that $a \geq b \geq c$.
If $b$ and $c$ are both negative, then so is $b+c$, which contradicts the given information. So there can be at most one negative value among the three, which with our ordering must be $c$.

In the case where $a, b$, and $c$ are all positive or $0$, then the positive (or zero) number $x=a+b+c$ is greater than or equal to $x / 3$.

Otherwise, since we have assumed that $c$ is the least of the three, $c$ is negative while $a$ and $b$ are not. Then $a+b+c=a+b-|c| \geq a$ since $b+c \geq 0$ tells us that $b \geq -c=|c|$. On the other hand $\frac{|a|+|b|+|c|}{3}=\frac{a+b-c}{3} \leq a$ since the average of three numbers is less than or equal to the greatest of the numbers. By transitivity we have $a+b+c \geq \frac{|a|+|b|+|c|}{3}$.

Alt Solution 2:
The case where $a, b$, and $c$ are all positive or zero can be handled as before. For the case where $a, b \geq 0$ and $c<0$, $a+b+c-\frac{|a|+|b|+|c|}{3}=a+b+c-\frac{a+b-c}{3}=\frac{2 a+2 b+4 c}{3}=\frac{2(a+b)+2(b+c)}{3} \geq 0$.