# A,b,c are given:

#### MathematicalPhysicist

Gold Member
and a+b+c>0 and abc>0
proove that abc>=a+b+c

#### bogdan

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...
[?]

#### MathematicalPhysicist

Gold Member
****, damn.
btw, bogdan, did these numbers were the first to pop up your head?

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#### STAii

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

I wonder if maybe this was supposed to be in the domain of all the positive integers?

#### MathematicalPhysicist

Gold Member
I wonder if maybe this was supposed to be in the domain of all the positive integers?
let's say it is, does it change the answer?

#### bogdan

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...

#### MathematicalPhysicist

Gold Member
do you have proof or what you have shown was your proof?

#### bogdan

(that's not true for integers equal to 1...in specific cases...)

#### MathematicalPhysicist

Gold Member
Originally posted by bogdan
(that's not true for integers equal to 1...in specific cases...)
thanks.

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