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

[astrololo]

Homework Statement

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

Homework Equations

I must prove this by the method of case.

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...

 

[SammyS]

Homework Statement

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

Homework Equations

I must prove this by the method of case.

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...

What are the cases you refer to ?

 

[astrololo]

What are the cases you refer to ?

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.

 

[astrololo]

What are the cases you refer to ?

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

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

 

[SammyS]

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.

I misread your post.

Case 1: x ≥ 0

If x ≥ 0, what is |x| ?
​

 

[astrololo]

I misread your post.

Case 1: x ≥ 0

If x ≥ 0, what is |x| ?
​

Then this means |x|=x

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

 

[SammyS]

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

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

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

 

[astrololo]

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

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 ?

 

[SammyS]

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 ?

Yes.

 

[astrololo]

Yes.

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

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

 

[Ray Vickson]

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

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

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 "$= 0$".