and a+b+c>0 and abc>0
proove that abc>=a+b+c
can someone please help?
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...
btw, bogdan, did these numbers were the first to pop up your head?
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?
let's say it is, does it change the answer?
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...
do you have proof or what you have shown was your proof?
That's the proof...read it carefuly...
(that's not true for integers equal to 1...in specific cases...)
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