0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Why do you have absolute values inside an absolute value? $\endgroup$ – dmtri Mar 12 at 19:17
2
$\begingroup$

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:

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.