# Showing that the mean of one number and its absolute val..

1. Oct 4, 2015

### astrololo

1. The problem statement, all variables and given/known data
I must prove that the mean of one number and its absolute value is superior or equal to 0 but inferior or equal to the absolute value of the number

2. Relevant equations
I must prove this by the method of case.

3. The attempt at a solution

I know that the first case is : x>=0 and the second x<0

But after that, I have no idea on how to do this....

2. Oct 4, 2015

### SammyS

Staff Emeritus
What are the cases you refer to ?

3. Oct 4, 2015

### astrololo

You mean you don't understand what the proof by case is ? I mean that to arrive to my result, I must begin x>=0 and the second case is x<0. After, I imagine there is some sort of addition that I Must do to obtain my final result.

4. Oct 4, 2015

### astrololo

Are you still here ? I think that I must do this : x<0 then absolute(x)=-x

and x>=0 then absolute(x)=x

5. Oct 4, 2015

### SammyS

Staff Emeritus

Case 1: x ≥ 0
If x ≥ 0, what is |x| ?

6. Oct 4, 2015

### astrololo

Then this means |x|=x

And if x<0 then this means |x|=-x

7. Oct 4, 2015

### SammyS

Staff Emeritus
Yes. That's the way to start each part.

8. Oct 4, 2015

### astrololo

Yeah, I was able to understand this. Here is what I did after :

|x|=x add x on each side

|x|+x=2x

divide by 2

(|x|+x)/2=x

Case 2 :

|x|=-x

|x|+x=x-x

|x|+x=0

Multiply by 1/2 by each side.

(|x|+x)/2=0

Is this correct ?

9. Oct 4, 2015

### SammyS

Staff Emeritus
Yes.

10. Oct 4, 2015

### astrololo

Ok, I guess that I must use my x<=0 inequality and replace the x.

(|x|+x)/2<=0

11. Oct 5, 2015

### Ray Vickson

No. You were asked to prove that
$$0 \leq \frac{x+|x|}{2} \leq |x|$$
Note that this is "$\geq 0$", not "$\leq 0$", but, of course, they are the same thing when you actually have "$= 0$".