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

Question

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

Answer 1

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.

Answer 2

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:

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

Answer 3

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.

Answer 4

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.