Is there a non-brute force way to determine $|\frac{1-x}{2}| + |\frac{1+x}{2}| \leq 1$? I am interested in determining all values for $x$ such that $|\frac{1-x}{2}| + |\frac{1+x}{2}| \leq 1$. I am aware that one can brute force a solution by trying all four combinations of signs, but is there a more elegant/quicker way to deal with such inequalities?
 A: Write $f(x) = |1-x| + |1+x|$. You are trying to find the values of $x$ such that $f(x) \leq 2$. Observe that
$$ 2 = |1-x+1+x| \leq |1-x|+|1+x| = f(x) $$
So, for every $x$, the only possible situation is that $f(x) = 2$. That is the solutions to $|1-x|+|1+x| = 2$ are the ones you are looking for. It happens at all $x \in [-1,1]$.
A: The left side of the inequality remains unchanged when replacing $x$ by $-x$, so without loss of generality we may assume $x \ge 0$.  Now $\frac{1+x}{2}$ is always positive so we need only consider two cases: $[0,1]$ and $[1,\infty)$.
A: Just use triangle inequality:
$$1=|\frac{1-x}{2} + \frac{1+x}{2}| <= |\frac{1-x}{2}| + |\frac{1+x}{2}| = 1$$
So we must have equality in the triangle and this can happen if both arguments have the same sign. Hence $1-x>=0$ and $1+x<=0$  or $1-x<=0$ and $1+x<=0$.
The first one gives $[-1,1]$. The second one has no solutions of course.
A: Note that $|x| = \max(x,-x)$ so
\begin{eqnarray}
|1+x| + |1-x| &=&\max(1+x,-1-x)+\max(1-x,x-1) \\
&=& \max(2,2x,-2x,-2) \\
&=& 2\max(1,|x|)
\end{eqnarray}
Then you need to solve $\max(1,|x|) \le 1$ which is seen to be $[-1,1]$.
