Slovenija 2008

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 $x\ge -1007$.