Prove the inequality

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mynamesmurph

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Any help?

A was easy enough and I can see the pattern. Perhaps, some help with part B?
 

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I reposted the image to show the problem in its entirety.

With regard to your remark, I'm not quite sure where to begin, maybe you could point me in the right direction?
 
I think my question is a more fundamental one... What is my end goal? Maybe explain the problem in a different way? I can rearrange the terms fairly easily.

√(ab) ≤ (a+b)/2


(ab) ≤ [(a+b)/2]^2

(ab) ≤ {[(a+b)]^2}/2

2ab ≤ [(a+b)]^2

2ab ≤ a^2+2ab+b^2

0 ≤ a^2-2ab+b^2

0 ≤ (a-b)^2

0 ≤ a-b

b ≤ a


Did I do that right? But again, I'm not quite sure what I'm supposed to be doing.
 
mynamesmurph said:
I think my question is a more fundamental one... What is my end goal? Maybe explain the problem in a different way? I can rearrange the terms fairly easily.

√(ab) ≤ (a+b)/2


(ab) ≤ [(a+b)/2]^2

(ab) ≤ {[(a+b)]^2}/2 why did you not square the 2?

2ab ≤ [(a+b)]^2

2ab ≤ a^2+2ab+b^2 when you have the same exact term on both sides of the equation when you subtract them from both sides they disappear as the subtraction results in 0

0 ≤ a^2-2ab+b^2 why on the lhs was 2ab-2ab=0 but on the rhs 2ab-2ab = -2ab??

0 ≤ (a-b)^2

0 ≤ a-b

b ≤ a


Did I do that right? But again, I'm not quite sure what I'm supposed to be doing.
Click to expand...
Of course something is wrong as you did not arrive at something that MUST be true. That is b does NOT have to less than = a. You want something at the end like 5<7 or a^2>=0. Actually you should have stopped at 0 ≤ (a-b)^2 since this always true for a,b non negative (actually it is always true!). The reason the proof is still wrong even if you removed the last two lines is because you made mistakes up to that (new) last line.

You need to understand what you are looking for (something always true under your constraints) so you know where to stop. Writing (and saying!) iff or <--> between each line might help.

One last point. a and b are symmetrical in this problem (you can interchange them and the problem is exactly the same--try it) so b<=a can't always be true because you can easily arrive at a<=b... .
 
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Sorry, I think I'm compounding the problem by shoddily reproducing my hand written notes into text. Those first errors cited are due to this.

Jomo said:
You need to understand what you are looking for (something always true under your constraints) so you know where to stop. Writing (and saying!) iff or <--> between each line might help.

One last point. a and b are symmetrical in this problem (you can interchange them and the problem is exactly the same--try it) so b<=a can't always be true because you can easily arrive at a<=b... .
Click to expand...

I will try to consider your statements further.
 
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mynamesmurph said:
Sorry, I think I'm compounding the problem by shoddily reproducing my hand written notes into text. Those first errors are cited are due to this.



Indeed, I need to know what I'm looking for. I still don't understand what my result is supposed to even be.
Click to expand...
You start off with the original statement which may or not be true and find something equivalent to that (which you do not know is always true) then find something equivalent to that statement (which you do not know is always true).... which is equivalent to a statement which you know is always true.Then the original statement is true since it is equivalent to a statement which you know is true

√(ab) ≤ (a+b)/2 is the same as
ab≤ [(a+b)/2]^2= [a^2 + 2ab + b^2]/4 is the same as
4ab≤ a^2 + 2ab + b^2 which is the same as
0≤ a^2 - 2ab + b^2 =(a-b)^2. This last statement, 0≤ (a-b)^2 is always true. So the original statement is √(ab) ≤ (a+b)/2

Basically if something is true (or false) will stay true (or false) if we multiply/divide both sides by the same value or add/subtract both sides by the same value or....

Let us know if you have any questions.
 
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√(ab) ≤ (a+b)/2


(ab) ≤ [(a+b)/2]^2

(ab) ≤ [(a+b)^2]/4

4ab ≤ (a+b)^2

4ab ≤ a^2+2ab+b^2

0 ≤ a^2-2ab+b^2

0 ≤ (a-b)^2



I removed the mistakes, (please tell me otherwise) but am not sure how to proceed. Within the context of this remark

Hint: Use the following property of inequalities:
If x and y are nonnegative, then the inequality x ≤ y is
equivalent to x2 ≤ y2



Am I supposed to show my statement is true, because I can demonstrate the first inequality is same as the inequality in the hint?





As an aside, maybe some clarification on taking the square root of both sides?

0 ≤ (a-b)^2
Taking the square root of both sides this should give me the absolute value of a-b on the right.

0 ≤ |a-b|

a-b is greater than or equal to zero away from zero, (which is all real numbers right?)

0 ≤ a-b or a-b ≥ 0

b ≤ a or a ≥ b

Earlier you mentioned this cannot be true. Maybe a few words about this?
 
mynamesmurph said:
√(ab) ≤ (a+b)/2


(ab) ≤ [(a+b)/2]^2

(ab) ≤ [(a+b)^2]/4

4ab ≤ (a+b)^2

4ab ≤ a^2+2ab+b^2

0 ≤ a^2-2ab+b^2

0 ≤ (a-b)^2



I removed the mistakes, (please tell me otherwise) but am not sure how to proceed. Within the context of this remark

Hint: Use the following property of inequalities:
If x and y are nonnegative, then the inequality x ≤ y is
equivalent to x2 ≤ y2

This hint says that if x>0, y>0 and x ≤ y then x2 ≤ y2
You did use this hint when you went from √(ab) ≤ (a+b)/2 to (ab) ≤ [(a+b)/2]^2



Am I supposed to show my statement is true, because I can demonstrate the first inequality is same as the inequality in the hint?
Yes, you are supposed to show that √(ab) ≤ (a+b)/2 is true. Well you know that it is exactly the same as (ie equivalent to) 0 ≤ (a-b)^2 which is true!





As an aside, maybe some clarification on taking the square root of both sides?

0 ≤ (a-b)^2
Taking the square root of both sides this should give me the absolute value of a-b on the right.

0 ≤ |a-b|

a-b is greater than or equal to zero away from zero, (which is all real numbers right?)

0 ≤ a-b or a-b ≥ 0 0< |3-7| but 3-7> 0 is NOT true

b ≤ a or a ≥ b

Earlier you mentioned this cannot be true. Maybe a few words about this?
Click to expand...
|a-b| either equals a-b or b-a depending on whether a<b or a>b.
So 0 ≤ |a-b| means 0≤ a-b iff a ≥ b OR 0 ≤ |a-b| means 0 ≤ b-a iff b ≥ a. So either a ≥ b OR b ≥ a. But this is always true!

I am glad that you investigated beyond 0 ≤ (a-b)^2 but the proof ends at that line (since we know for sure that if a ≥ 0 and b ≥ 0 then 0 ≤ (a-b)^2).
 
OK, I think I got that.


Moving along to part C)

Assuming a=b and it nonnegative

√(ab) = a+b/2

√(aa) = a+a/2

√(a^2) = 2a/2

|(a)| = a


Right?
 
and D) Assuming they are now equal instead of less than or equal to, and non-negative, show a=b

(using the same steps we get to this)

0 = |(a-b)|


Using the def of absolute value

|(a-b)| = -(a-b) if (a-b) < 0

So, when (a-b) < 0
0 = -(a-b)
and
a=b


Def of absolute value
|(a-b)| = (a-b) if (a-b) ≥ 0


So, when (a-b) ≥ 0
0 = (a-b)
and
b=a



Right? Did I make a mistake, or mess up some logical part, or too wordy?
 
mynamesmurph said:
and D) Assuming they are now equal instead of less than or equal to, and non-negative, show a=b

(using the same steps we get to this)

0 = |(a-b)|


Using the def of absolute value

|(a-b)| = -(a-b) if (a-b) < 0

So, when (a-b) < 0
0 = -(a-b)
and
a=b


Def of absolute value
|(a-b)| = (a-b) if (a-b) ≥ 0


So, when (a-b) ≥ 0
0 = (a-b)
and
b=a



Right? Did I make a mistake, or mess up some logical part, or too wordy?
Click to expand...
Maybe it is right but it was too painful for me to follow.

We know that there is exactly one value that we can take the absolute value and get 0, and that value is 0, that is if |x|=0 then x=0.
So if 0 = |(a-b)|, then a-b = 0 and a=b. Done
 
Jomo said:
Maybe it is right but it was too painful for me to follow.

We know that there is exactly one value that we can take the absolute value and get 0, and that value is 0, that is if |x|=0 then x=0.
So if 0 = |(a-b)|, then a-b = 0 and a=b. Done
Click to expand...

Haha, sorry about that. Thanks for the help.
 
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