# How can this be proven? (maths)

**URL:** <https://boards.straightdope.com/t/how-can-this-be-proven-maths/292505>\
**Category:** Factual Questions\
**Created:** [March 1, 2005, 11:06pm UTC](https://boards.straightdope.com/t/how-can-this-be-proven-maths/292505 "2005-03-01T23:06:42Z")\
**Posts on this page:** 9\
**Page:** 2

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**Author:** ![emarkp](https://avatars.discourse-cdn.com/v4/letter/e/3be4f8/32.png) [@emarkp](https://boards.straightdope.com/u/emarkp)\
**Post date:** [March 2, 2005, 8:43pm UTC](https://boards.straightdope.com/t/how-can-this-be-proven-maths/292505/21 "2005-03-02T20:43:50Z")

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The answer is to use logarithms.

Since k,x are both positive, all the quantities are positive, and so their logarithms will have the same ordering as the quantities themselves. Given that:

Prove: (1+x)[sup]k[/sup] \<= 1 + kx

is equivalent to proving:  
log((1+x)[sup]k[/sup]) \<= log(1 + kx)

is equivalent to proving:  
k log(1+x) \<= log(1 + kx)

Now since 0 \< k \<= 1, and x \> 0

k log(1+x) \<= log(1+x)  
(1+x) \<= (1+kx)

and hence:

k log(1+x) \<= log(1+x) \<= log(1+kx)

and you’re done.

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**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\
**Post date:** [March 2, 2005, 8:46pm UTC](https://boards.straightdope.com/t/how-can-this-be-proven-maths/292505/22 "2005-03-02T20:46:13Z")

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If k is in (0, 1], then 1 + x \> 1 + kx.

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**Author:** ![emarkp](https://avatars.discourse-cdn.com/v4/letter/e/3be4f8/32.png) [@emarkp](https://boards.straightdope.com/u/emarkp)\
**Post date:** [March 2, 2005, 8:51pm UTC](https://boards.straightdope.com/t/how-can-this-be-proven-maths/292505/23 "2005-03-02T20:51:17Z")

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oops. Sigh. My first thought was derivatives too. Should have stuck with it.

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**Author:** ![SCSimmons](https://avatars.discourse-cdn.com/v4/letter/s/e495f1/32.png) [@SCSimmons](https://boards.straightdope.com/u/SCSimmons)\
**Post date:** [March 3, 2005, 12:13am UTC](https://boards.straightdope.com/t/how-can-this-be-proven-maths/292505/24 "2005-03-03T00:13:46Z")

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If k is restricted to integers, this is very easily proved by induction on k …

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**Author:** ![Cabbage](https://avatars.discourse-cdn.com/v4/letter/c/f07891/32.png) [@Cabbage](https://boards.straightdope.com/u/Cabbage)\
**Post date:** [March 3, 2005, 12:45am UTC](https://boards.straightdope.com/t/how-can-this-be-proven-maths/292505/25 "2005-03-03T00:45:20Z")

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> [@SCSimmons](#):
>
> If k is restricted to integers, this is very easily proved by induction on k …

I think you misread the statement. If k is an integer greater than 1, the inequality is false (it’s reversed).

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**Author:** ![Tyrrell\_McAllister](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/tyrrell_mcallister/32/16772_2.png) [@Tyrrell\_McAllister](https://boards.straightdope.com/u/Tyrrell_McAllister)\
**Post date:** [March 3, 2005, 9:56am UTC](https://boards.straightdope.com/t/how-can-this-be-proven-maths/292505/26 "2005-03-03T09:56:46Z")

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> [@Cabbage](#):
>
> I think you misread the statement. If k is an integer greater than 1, the inequality is false (it’s reversed).

Well, in this case, “_k_ restricted to integers” means _k_ = 1, and that case is very easy to prove by induction ;).

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**Author:** ![SCSimmons](https://avatars.discourse-cdn.com/v4/letter/s/e495f1/32.png) [@SCSimmons](https://boards.straightdope.com/u/SCSimmons)\
**Post date:** [March 3, 2005, 2:41pm UTC](https://boards.straightdope.com/t/how-can-this-be-proven-maths/292505/27 "2005-03-03T14:41:37Z")

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> [@Cabbage](#):
>
> I think you misread the statement. If k is an integer greater than 1, the inequality is false (it’s reversed).

You’re right, I did. But I successfully proved the reversed inequality for k\>=1. 🙂

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**Author:** ![Tyrrell\_McAllister](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/tyrrell_mcallister/32/16772_2.png) [@Tyrrell\_McAllister](https://boards.straightdope.com/u/Tyrrell_McAllister)\
**Post date:** [March 3, 2005, 4:24pm UTC](https://boards.straightdope.com/t/how-can-this-be-proven-maths/292505/28 "2005-03-03T16:24:47Z")

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Here’s a solution that does not use derivatives. It only depends on the fact that (1) for fixed α \> 0, _y_[sup]&nbsp;α[/sup] increases (strictly) monotonically as a function of _y_ on [1, ∞), and (2) the product of two monotonically increasing functions is monotonically increasing.

The problem is trivial if _k_ = 1, so fix _k_ such that 0 \< _k_ \< 1. Making the substitution _y_ = 1 + _x_ and rearranging, the problem is equivalent to showing that if _y_ \> 1, then  
_k_ _y_ − _y_[sup]&nbsp;_k_[/sup] ≥ _k_ − 1.

Observe that the right hand side of this inequality is a constant (since _k_ is fixed) and that the inequality is satisfied with equality when _y_ = 1. Hence, it suffices to show that _k_ _y_ − _y_[sup]&nbsp;_k_[/sup] increases monotonically as a function of _y_ on [1, ∞).

Indeed, we have that

_k_ _y_ − _y_[sup]&nbsp;_k_[/sup] = (_y_[sup]1 − _k_[/sup] − _k_[sup]&nbsp;−1[/sup]) _k_ _y_[sup]&nbsp;_k_[/sup],

so _k_ _y_ − _y_[sup]&nbsp;_k_[/sup] is the product of two monotonically increasing functions. Hence, _k_ _y_ − _y_[sup]&nbsp;_k_[/sup] is itself monotonically increasing, so the claim is proved.

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**Author:** ![Mathochist](https://avatars.discourse-cdn.com/v4/letter/m/c89c15/32.png) [@Mathochist](https://boards.straightdope.com/u/Mathochist)\
**Post date:** [March 3, 2005, 5:13pm UTC](https://boards.straightdope.com/t/how-can-this-be-proven-maths/292505/29 "2005-03-03T17:13:54Z")

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The logarithm proof is pretty good, but there’s a variant on the derivative proof: expand the Taylor series about the origin and use the remainder formula after a few terms.

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