- #1
MathematicalPhysicist
Gold Member
- 4,699
- 371
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?
- Thread starter MathematicalPhysicist
- Start date
In summary, the conversation discusses how to prove the inequality abc>=a+b+c for positive numbers. The participants use examples and mathematical logic to show that the inequality holds true for all positive integers. Ultimately, it is determined that the inequality is only false in specific cases and generally holds true.
- #2
bogdan
- 191
- 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
MathematicalPhysicist
Gold Member
- 4,699
- 371
****, 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
- 333
- 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
- 502
- 1
I wonder if maybe this was supposed to be in the domain of all the positive integers?
- #6
MathematicalPhysicist
Gold Member
- 4,699
- 371
let's say it is, does it change the answer?
- #7
bogdan
- 191
- 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
MathematicalPhysicist
Gold Member
- 4,699
- 371
do you have proof or what you have shown was your proof?
- #9
bogdan
- 191
- 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
MathematicalPhysicist
Gold Member
- 4,699
- 371
thanks.Originally posted by bogdan
That's the proof...read it carefuly...
(that's not true for integers equal to 1...in specific cases...)
1. What does "A, b, c are given" mean?
When a statement says "A, b, c are given", it means that the values of A, b, and c have been previously determined or provided. These values are known and can be used in calculations or experiments.
2. How do you determine the values of A, b, and c?
The values of A, b, and c can be determined through various methods, depending on the context. For example, in a mathematical equation, A, b, and c may be constants that are already known or can be solved for. In a scientific experiment, A, b, and c may be measured or controlled variables.
3. Why is it important to know that A, b, and c are given?
Knowing that A, b, and c are given is important because it helps us to understand the context and limitations of a problem or experiment. It also allows us to accurately interpret and use the results and make informed conclusions.
4. Can A, b, and c be changed or manipulated?
In most cases, the values of A, b, and c are considered fixed and cannot be changed or manipulated. However, there may be situations where these values can be adjusted or controlled, such as in a controlled scientific experiment.
5. Are there alternative notations for "A, b, c are given"?
Yes, there are alternative notations for indicating that A, b, and c are given. Some examples include "A, b, and c are known", "A, b, and c are given values", or simply "given A, b, c". The specific notation used may vary depending on the context and discipline.
Similar threads
-
General Math
-
General Math
-
General Math
-
General Math
-
General Math
-
General Math
-
General Math
-
General Math
-
General Math