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Binomial Theorem-Bernoulli Inequality

I was looking at the proof of Bernoulli Inequality using binomial theorem on Wikipedia. There is some parts within the proof that is not clear to me that is; given 0 < y ≤ 1

(Line 1)
Notice that

I am okay up to this point, but then it says

Therefore, we can see that each binomial term

is multiplied by a factor

, and that will make each term smaller than the term before.
For that reason,

(Line 2)

But it doesn't make sense since if there was a second term on line 2 (the y^3 term) then it would clearly be (-1)*rC3*y^3 which is negative but then a third term (y^4 term) is rC4*y^4 which is positive so then the third term is not smaller than the second term as the proof suggests.

If by "that will make each term smaller than the term before" it is implying that each term on line 2 will be smaller than its correspondent on line 1, then it still does not make sense since

(-1)*rC3 is less than or equal to (-1)*rC3*y^3 for 0 < y ≤ 1

Can somebody help with this and explain what I am missing...
It may be helpful if i share the link of proof (https://en.wikipedia.org/wiki/Bernoulli%27s_inequality) so you can follow everything...
or just clarify why line 2 greater than or equal to zero given line 1...

Reply 1

username4240292

They referring to unsigned terms I think (i.e. the magnitude); it's not written too well, but I think they're saying:

[EDIT: wrong - see below].

(edited 5 years ago)

Reply 2

ghostwalker

Original post by ftfy

Every term containing an even power of

y

\ge

than the next term containing an odd power of

y

. So their difference is non-negative i.e. we're grouping terms whose difference is non-negative. So the whole lot after

+ \cdots

is non-negative.

Alas, that isn't true. Take r=8, y= 0.75 for example, then the magnitude (absolute value) of the y^3 term is greater than that of the y^2, and that pair come out negative.

Edit:

And

\displaystyle \binom{r}{n} \ge \binom{r}{m}

for

m\ge n

isn't always true

(edited 5 years ago)

Reply 3

ghostwalker

Original post by Merdan

or just clarify why line 2 greater than or equal to zero given line 1...

I've looked at it for a while, and I can't see the justification.

Come across very few references on the 'net that look at negative x with the binomial, and none of them provide any further justification.

Reply 4

username4240292

Original post by ghostwalker

Alas, that isn't true. Take r=8, y= 0.75 for example, then the magnitude (absolute value) of the y^3 term is greater than that of the y^2, and that pair come out negative.

Edit:

And

\displaystyle \binom{r}{n} \ge \binom{r}{m}

for

m\ge n

isn't always true

Yep, it isn't true. It's likely that the person writing the article used the same wrong binomial inequality as me.

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