Minimizing Sum Of Absolutes

Relevant wiki: Absolute Value Inequalities - 3 or more Linear Terms

Absolute value functions can be written as piecewise functions:

$|x-a|=\begin{cases}x&x\geq 0\\ -x&x<0\end{cases}$

We can rewrite each absolute function above as piecewise functions:

$|x+2|=\begin{cases}x+2&x\geq -2\\ -x-2&x<-2\end{cases}$

$|x+3|=\begin{cases}x+3&x\geq -3\\ -x-3&x<-3\end{cases}$

$|x+4|=\begin{cases}x+4&x\geq -4\\ -x-4&x<-4\end{cases}$

Combine all 3 of them, and we have 4 different cases:

$|x+2|+|x+3|+|x+4|=\begin{cases}3x+9&x\geq-2\\ x+5&-3 \leq x < -2\\ -x-1&-4 \leq x < -3\\ -3x-9& x < -4\end{cases}$

The range for each case is:

$\begin{cases}3x+9&x\geq-2&\implies[3,\infty)\\ x+5&-3 \leq x < -2&\implies[2,3)\\ -x-1&-4 \leq x < -3&\implies(2,3]\\ -3x-9& x < -4&\implies(3,\infty)\end{cases}$

Therefore, the range of $|x+2|+|x+3|+|x+4|$ is $[2,\infty)$

The minimum value of the expression is $\boxed{2}$ , which occurs at $x=-3$

The minimum value of |x+2| + |x+4| is 2, at all values between -2 and -4. Since the minimum of |x+3| is 0, and that minimum is at -3, which is within the minimum range for the other two, the total minimum is 2.