# inequality $x > \left | \left | x \right | + \left | x-3 \right |+\left | x-5 \right |\right |+7$

I need help in this exercise because I got stuck in this inequality which I have to show that it has no solution, could someone give me an idea of ​​how to continue or finish it, please. $$x > \left | \left | x \right | + \left | x-3 \right |+\left | x-5 \right |\right |+7$$

• Why do you have absolute values inside an absolute value? – dmtri Mar 12 at 19:17

First, note that the outer absolute value is unnecessary since all your values are positive, so your equation reduces to

$$x > |x| + |x - 3| + |x - 5| + 7$$

From this, we can clearly see that the equation is impossible.

This is because:

$$|x| + |x - 3| + |x - 5| + 7 > |x| + 7 > x$$

First observe $$x>7. \tag{1}$$ Then $$|x|=x,|x-3|=x-3,|x-5|=x-5$$ and hence the inequality becomes $$x>3x-1$$ which implies $$x<\frac12. \tag{2}$$ Note that (1) and (2) are contradictory with each other. So the inequality has no solution.

The RHS is a piecewise linear function, which is continuous. It suffices to check inequality at the angular points and outside.

• $$x\le0\to x>|-x-(x-3)-(x-5)|+7$$ is certainly impossible,

• $$x=3\to 3>|3+0+2|+7$$ does not hold,

• $$x\ge5\to x>|x+x+2+x-5|+7$$ is certainly impossible.

If you just logically look at it, it’s obviously not always true, thus you can prove it is not true by giving a counter example.