- #1
- 4,662
- 372
and a+b+c>0 and abc>0
proove that abc>=a+b+c
can someone please help?
proove that abc>=a+b+c
can someone please help?
Is ABC Greater Than or Equal to A+B+C?
- Context: Undergrad
- Thread starter MathematicalPhysicist
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Discussion Overview
The discussion revolves around the inequality abc ≥ a + b + c, under the conditions that a, b, and c are positive numbers. Participants explore various cases, challenge the validity of the inequality, and consider whether it holds true within specific domains, such as positive integers.
Discussion Character
- Debate/contested
- Mathematical reasoning
Main Points Raised
- One participant asserts that if a, b, and c are positive and a + b + c > 0, then abc should be greater than or equal to a + b + c.
- Another participant provides a counterexample with a = 0.1, b = 0.1, c = 0.1, showing that abc = 0.001 is less than a + b + c = 0.3, suggesting the inequality does not hold in this case.
- A participant elaborates on the reasoning behind choosing small values for a, b, and c to disprove the inequality, discussing the behavior of products of positive numbers.
- There is a suggestion that the inequality might only apply to positive integers, prompting further discussion on whether this changes the validity of the inequality.
- Another participant claims that for positive integers, the inequality holds true, providing a reasoning based on the relationships between the variables.
- A request for proof is made, and a participant claims that their previous explanation serves as proof, while also noting exceptions for specific cases involving the integer 1.
Areas of Agreement / Disagreement
Participants express disagreement regarding the validity of the inequality under different conditions. Some argue it holds for positive integers, while others provide counterexamples suggesting it does not hold for all positive numbers.
Contextual Notes
The discussion highlights the dependence on the definitions and domains of the variables involved, particularly whether they are restricted to positive integers or include all positive real numbers.
- #2
bogdan
- 188
- 0
I'm afraid it's not correct...
if a=0.1,b=0.1,c=0.1 then abc=0.001 and a+b+c=0.3, 0.001<0.3...
[?]
if a=0.1,b=0.1,c=0.1 then abc=0.001 and a+b+c=0.3, 0.001<0.3...
[?]
- #3
- 4,662
- 372
****, damn.
btw, bogdan, did these numbers were the first to pop up your head?
btw, bogdan, did these numbers were the first to pop up your head?
Last edited:
- #4
STAii
- 327
- 1
It is not really hard to get those numbers.
All you need to know is think for a moment ..
If you have a positive number (call it X), and you multiply it but another positive number (say Y), you can get one of the following results :
XY > X
XY < X
XY = X
Now, if XY = X then Y = 1, obviously this is the changing point between XY > X and XY < X
Try numbers bigger than 1 for Y, say, Y = 2, then XY = 2X, therefore XY > X (remember that X is positive).
Now, try numbers smaller than 1 (and more than 0, remember Y is positive), say, Y = 0.1, then you get XY = X/10, and therefore XY < X
bogdan was trying to disproove the original inequality, logically if he gets a sinlge case that gives a wrong answer in the inquality, then the inquality is wrong.
The inequality says that abc will be bigger or equal to a+b+c,
So, to disproove this try to get smaller value of abc than expected (since this MIGHT turn the inequality wrong, by making left side smaller).
Since values smaller than 1 for a,b,c will make abc smaller and smaller, bogdan have chosen 0.1 for all of them.
I see infinite number of cases that proove the inequality wrong.
Try those :
1- a=0.01, b=0.1, c=0.1
2- a=1, b=1, c=1
3- a=1.1, b=0.0001, c=0.0002
All you need to know is think for a moment ..
If you have a positive number (call it X), and you multiply it but another positive number (say Y), you can get one of the following results :
XY > X
XY < X
XY = X
Now, if XY = X then Y = 1, obviously this is the changing point between XY > X and XY < X
Try numbers bigger than 1 for Y, say, Y = 2, then XY = 2X, therefore XY > X (remember that X is positive).
Now, try numbers smaller than 1 (and more than 0, remember Y is positive), say, Y = 0.1, then you get XY = X/10, and therefore XY < X
bogdan was trying to disproove the original inequality, logically if he gets a sinlge case that gives a wrong answer in the inquality, then the inquality is wrong.
The inequality says that abc will be bigger or equal to a+b+c,
So, to disproove this try to get smaller value of abc than expected (since this MIGHT turn the inequality wrong, by making left side smaller).
Since values smaller than 1 for a,b,c will make abc smaller and smaller, bogdan have chosen 0.1 for all of them.
I see infinite number of cases that proove the inequality wrong.
Try those :
1- a=0.01, b=0.1, c=0.1
2- a=1, b=1, c=1
3- a=1.1, b=0.0001, c=0.0002
- #5
Brad_Ad23
- 497
- 1
I wonder if maybe this was supposed to be in the domain of all the positive integers?
- #6
- 4,662
- 372
let's say it is, does it change the answer?
- #7
bogdan
- 188
- 0
Yes...for positive integers it is true...
let's say 2<=a<=b<=c...it doesn't "particularize"...
we have abc>=2*(bc)>=4*c=c+c+c+c>a+b+c...
let's say 2<=a<=b<=c...it doesn't "particularize"...
we have abc>=2*(bc)>=4*c=c+c+c+c>a+b+c...
- #8
- 4,662
- 372
do you have proof or what you have shown was your proof?
- #9
bogdan
- 188
- 0
That's the proof...read it carefuly...
(that's not true for integers equal to 1...in specific cases...)
(that's not true for integers equal to 1...in specific cases...)
- #10
- 4,662
- 372
thanks.Originally posted by bogdan
That's the proof...read it carefuly...
(that's not true for integers equal to 1...in specific cases...)
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