An Inequality (1 Viewer)

vds700 · Oct 5, 2008

If a,b are positive, unequal numbers, prove that

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thanks

tommykins · Oct 5, 2008

Abit of a controversial proof but here we go -

a>0
b>0
a-b > 0
OR
b-a > 0

.'. a>b or b>a
a^a > b^a and b^b>a^b

multiplication - you get a^a.b^b > b^a.a^b

conics2008 · Oct 5, 2008

thats some crazy shit u go there mate. im stickin to normal a>b

Undermyskin · Oct 5, 2008

This is how I do it because it makes sense to me. Hope it is right. Ala:

Let a>b
--> a = b + x

a^a = a ^ (b+x)
b^b = b ^ (a-x)

--> (a^b)(b^a)(a/b)^x (*)

since a>b, a/b > 1 , hence (a/b) ^ x > 1^x=1

Thus, (*) > (a^b)(b^a)

QED.

Undermyskin · Oct 5, 2008

tommykins said:
Abit of a controversial proof but here we go -

a>0
b>0
a-b > 0
OR
b-a > 0

.'. a>b or b>a
a^a > b^a and b^b>a^b

multiplication - you get a^a.b^b > b^a.b^a

I really don't get this. What did you multiply with, for human's sake?

You are having an 'or' so you can't make it an 'and' that easily.

tommykins · Oct 5, 2008

Undermyskin said:
I really don't get this. What did you multiply with, for human's sake?

You are having an 'or' so you can't make it an 'and' that easily.

Two different cases, and I accidentally typed in the letters wrong, these a's and b's confused me.

gh0stface · Oct 5, 2008

huh^ wtf.

your not answerin the question now,

tommykins · Oct 5, 2008

gh0stface said:
huh^ wtf.

your not answerin the question now,

...can you read?

gh0stface · Oct 5, 2008

^ oops makes sense. now

Timothy.Siu · Oct 5, 2008

i'm not sure if my working works coz i've never done proving inequalities but i'll try
a^a.b^b>a^b.b^a
a^(a-b) > b^(a-b)
if a>b then a-b>0 therefore a^(a-b) > b^(a-b)
if b>a then a-b<0 therefore a^(a-b) > b^(a-b)

Trebla · Oct 5, 2008

If we assume a > b
a^{a – b} > b^{a – b}
a^a/a^b > b^a/b^b
a^ab^b > b^aa^b
Now if b > a instead, then it’s the same thing but we swap a and b and we end up with the same inequality…so it holds regardless if whether a or b is greater.

tommykins · Oct 5, 2008

Trebla said:
If we assume a > b
a^{a – b} > b^{a – b}
a^a/a^b > b^a/b^b
a^ab^b > b^aa^b
Now if b > a instead, then it’s the same thing but we swap a with b and end up with the same inequality…so it holds regardless if whether a or b is greater.

Thank you for that, is my proof valid however?

Timothy.Siu · Oct 5, 2008

Trebla said:
If we assume a > b
a^{a – b} > b^{a – b}
a^a/a^b > b^a/b^b
a^ab^b > b^aa^b
Now if b > a instead, then it’s the same thing but we swap a with b and end up with the same inequality…so it holds regardless if whether a or b is greater.

lol thats wat i posted

except urs is better i guess, correct format

Undermyskin · Oct 5, 2008

tommykins, I really think you should reread your proof carefully because I can't find any connection between your last two lines! You confirm that you are working on two cases at the same time but then how can you assume the data of one case and apply to the other?

Timothy.Siu · Oct 5, 2008

yeah i dont think u can just change an "or" into an and

tommykins · Oct 5, 2008

yeah, its controversial as i said.

gh0stface · Oct 5, 2008

it just means that since you proved a^a > b^a and b^b>a^b

then if the two greater numbers are a^a and b^b multiplied together, it will still be greater the two smaller numbers multiplied b^a and a^b
i fink^ lol

tommykins · Oct 5, 2008

what i had in mind was like

a+b > 2sqrtab
b+c > 2sqrtbc
a+c > 2sqrtac

thus (a+b)(b+c)(a+c) > 8abc

Timothy.Siu · Oct 6, 2008

u were probably thinking too complicated maybe, =)

vds700 · Oct 6, 2008

Thanks everyine for their contribution

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