Problem 1145

We consider two cases depending on the sign of $2-x$.

If $x \ge 2$, then we have $|2-x| = -(2-x)$ and the inequality can be rewritten as $||x-2-x|-8| \le 2008$ or, equivalently, $6 \le 2008$. So for all real numbers $x \ge 2$ the inequality holds.

Now, let $x < 2$. In this case we have
$$||2 - 2x| - 8| \le 2008.$$
Let us consider two further cases depending on the sign of $2-2x$.

If $x \ge 1$, we have $|2-2x| = -(2-2x)$ and so $|2x-10| \le 2008$. Obviously, this inequality holds for $1 \le x < 2$.

Finally, only the case $x < 1$ remains. The condition implies $|-2x+2-8| \le 2008$ or $|2x+6| \le 2008$. This is equivalent to
$$-2008 \le 2x + 6 \le 2008.$$
Since $x < 1$ we have $2x + 6 < 2 + 6 = 8$, so the right inequality holds. The left inequality implies $-2014 \le 2x$ or $-1007 \le x$. From this we derive the condition $-1007 \le x < 1$.

Hence, the inequality holds for all $x \ge -1007$.