An Inequality (1 Viewer)

[FreakFace]

Consider
b^(b-a) + a^(b-a) > 0

=> Multiplying both sides by b^(b-a) -a^(b-a) we get

[b^(b-a) + a^(b-a) ][b^(b-a) -a^(b-a)] > 0

=> Difference of two squares

b^(2(b-a)) - a^(2(b-a)) > 0
b^(2(b-a)) > a^(2(b-a)

=>take square root

b^(b-a) > a^(b-a)

=>change negative power into fraction

b^b / b^a > a^b / a^a

=> Multiply both sides by by a^a . b^a

a^a . b^b > a^b . b^a



i think this works out nicely - using reverse method

 

[mick135]

Consider
b^(b-a) + a^(b-a) > 0

=> Multiplying both sides by b^(b-a) -a^(b-a) we get

[b^(b-a) + a^(b-a) ][b^(b-a) -a^(b-a)] > 0

=> Difference of two squares

b^(2(b-a)) - a^(2(b-a)) > 0
b^(2(b-a)) > a^(2(b-a)

=>take square root

b^(b-a) > a^(b-a)

=>change negative power into fraction

b^b / b^a > a^b / a^a

=> Multiply both sides by by a^a . b^a

a^a . b^b > a^b . b^a



i think this works out nicely - using reverse method

wow - i got confused reading that the first time, and understood it less the 2nd.
after 2 panadols i finally get it now XD - THANKS PANADOL - i mean freakface haha

 

[adnan91]

for you ppl on the front page or so u cant use the RHS in ur proof

 

[Timothy.Siu]

for you ppl on the front page or so u cant use the RHS in ur proof

wat do u mean

 

[lyounamu]

for you ppl on the front page or so u cant use the RHS in ur proof

Yes, you can. It's a "prove" question, not "show".

 

[FreakFace]

i dont see how you can assume a>b , as you're only given that a and b are real positive integers.

 

[Iruka]

You write "without loss of generality..." (or "wlog" if you want to be really terse) at the beginning of the proof.

It means that whichever of the two positive constants is the larger, we shall relabel it a, and the other b.

 

[Trebla]

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.

This is probably the simplest way to do it since a and b are not equal, so one is greater than the other. The proof works for BOTH possibilities of a being greater than b and vice versa (i.e. without loss of generality). You can also start with a < b to prove the other case, and you end up with same inequality.

 

[FreakFace]

is my method correct though ...

im not rlly sure about multiplying both sides by
b^(b-a) -a^(b-a)

because if a>b then that will be a negative number and flip the sign, but as someone said i think it may result in the same inequality.

 

[Timothy.Siu]

i dont see how you can assume a>b , as you're only given that a and b are real positive integers.

well one of the proofs assumed both so yeah.

 

[FreakFace]

here maybe its more understandable in this format