Need a proof (real analysis)

[CliveStaples]

Can anyone prove the following:

Let g be a real number such that 0 <= g <= 1. Let m,n be integers.

Conjecture 1: a^g + b^g - 1 >= (a+b-1)^g

Conjecture 2: if a >= b, then b^g - (b-1)^g >= a^g - (a-1)^g

[Categories+Sheaves]

You can do the first conjecture in one line by multiplying both sides by (1/3)g, yielding
(a/3)g + (b/3)g +(1/3)g >= (a/3 + b/3 + 1/3)g
Which is true by the concavity of f(x) = xg for g in [0,1]

#2 is simple if you interpret each side as an integral of xg's derivative (which is monotonically decreasing, etc.).

Anyway, that's the easy route. Since you've specified that this is a simple real analysis question, were you looking for a purely algebraic argument? (because it feels like there should be one...)


afterthought: you meant "let a,b be integers", right?

[CliveStaples]

You can do the first conjecture in one line by multiplying both sides by (1/3)g, yielding
(a/3)g + (b/3)g +(1/3)g >= (a/3 + b/3 + 1/3)g
Which is true by the concavity of f(x) = xg for g in [0,1]

You've improperly switched some negatives to positives there, I think.

#2 is simple if you interpret each side as an integral of xg's derivative (which is monotonically decreasing, etc.).

Anyway, that's the easy route. Since you've specified that this is a simple real analysis question, were you looking for a purely algebraic argument? (because it feels like there should be one...)

Any proof will do, it's a useful lemma for my girlfriend's research.

afterthought: you meant "let a,b be integers", right?

Yeah, I messed with the formatting and forgot to change my variable names.

[Categories+Sheaves]

[quote='Categories+Sheaves' pid='317718' dateline='1343773102']
You can do the first conjecture in one line by multiplying both sides by (1/3)g, yielding
(a/3)g + (b/3)g +(1/3)g >= (a/3 + b/3 + 1/3)g

You've improperly switched some negatives to positives there, I think.

Ah. Indeed I did.
So that gives us (a-1)g + (b)g >= (a + b - 1)g as long as a > 1 (negatives makes this harder :/ ).
By conjecture #2...
1-0 >= ag - (a-1)g and so (a-1)g >= ag - 1
So just using concavity isn't enough...

Any proof will do, it's a useful lemma for my girlfriend's research.

Because there's nothing more interesting than another person's question: what's the problem/literature you're looking at?

[CliveStaples]

[quote='Categories+Sheaves' pid='317718' dateline='1343773102']
You can do the first conjecture in one line by multiplying both sides by (1/3)g, yielding
(a/3)g + (b/3)g +(1/3)g >= (a/3 + b/3 + 1/3)g

You've improperly switched some negatives to positives there, I think.

Ah. Indeed I did.
So that gives us (a-1)g + (b)g >= (a + b - 1)g as long as a > 1 (negatives makes this harder :/ ).
By conjecture #2...
1-0 >= ag - (a-1)g and so (a-1)g >= ag - 1
So just using concavity isn't enough...

Any proof will do, it's a useful lemma for my girlfriend's research.

Because there's nothing more interesting than another person's question: what's the problem/literature you're looking at?

It's to prove a lower bound for a function on a network that gives the 'energy' of the network.

Conjecture 2 can be used to prove conjecture 1 through induction.

[Categories+Sheaves]

Conjecture 2 can be used to prove conjecture 1 through induction.

Aww. Well, I just threw together an analytic proof using a,b>=1
WOLOG spz. b>=a>=1
Noting that Int(gx1-g on [0,1]) = 1 <--everyday integration
(a+b-1)g = (b)g + Int(gx1-g on [b,b+a-1])
(a)g = Int(gx1-g on [0,a]) = 1 + Int(gx1-g on [1,a])
And then Int(gx1-g on [1,a]) >= Int(gx1-g on [b,b+a-1]) for the usual reason, and the result follows.

It's to prove a lower bound for a function on a network that gives the 'energy' of the network.

Does the function have a name?
-----edited because CliveStaples has a sharper eye than I do-----

Also: shorter proof.
(a+b-1)g = (b)g + Int(gx1-g on [b,b+a-1]) =< (b)g + Int(gx1-g on [1,a]) = (b)g + (a)g -1
With the middle inequality coming form concavity.

[CliveStaples]

Noting that Int(xg on [0,1]) = 1

Int(x^g) on [0,1] = 1/(g+1)

[Categories+Sheaves]

Noting that Int(xg on [0,1]) = 1

Int(x^g) on [0,1] = 1/(g+1)

Pardon my spelling, meant the derivative

[CliveStaples]

Hey, thanks for the work! We were stuck trying algebra and couldn't figure it out; using integrals is pretty slick. I haven't actually written many proofs using integrals--except maybe in differential equations.