A Generalization of the Bernoulli Inequality
by djmathman, Oct 7, 2023, 6:44 PM
Here's a nice generalization of Bernoulli's Inequality I found the other day. This result was originally recorded in [1] (possibly earlier, I honestly can't tell!) and is, in my opinion, less known than it should be.
Recall the statement of Bernoulli's Inequality:
whenever 
and 
(with the reverse inequality occuring when 
). In certain special cases, this is easy to see; for example, when 
and 
,
![\[
(1+x)^3 = 1 + 3x + 3x^2 + x^3 \geq 1 + 3x.
\]](//latex.artofproblemsolving.com/2/e/0/2e02f384a742ae503384c9914978968032b61ff8.png)
However, such a naive process does not necessarily work when 
is not an integer. For example, consider the case 
, which has Taylor expansion
![\[
\sqrt{1+x} = 1 + \frac x2 - \frac18x^2 + \frac{1}{16}x^3 - \frac{5}{128}x^4 + \frac{7}{256}x^5 - \cdots.
\]](//latex.artofproblemsolving.com/7/a/e/7ae61d9f990ed9a0094bfbe7ed57c7c45ef91441.png)
Here the Taylor expansion consists of both positive and negative coefficients, so more subtle analysis is needed to make the above approach work. Nevertheless, Bernoulli's Inequality does say that 
for all 
.
At this point, we raise a natural question: does a "quadratic" version of Bernoulli's Inequality
![\[
(1+x)^r \geq 1 + rx + \frac{r(r-1)}2x^2\qquad(1)
\]](//latex.artofproblemsolving.com/0/b/a/0bab5730a1150914079f08711eeb1858f573488a.png)
hold in general? What about higher approximations? It turns out the general criterion is surprisingly simple.
Theorem ([1]). Let
be an arbitrary real number, and consider a hypothetical inequality of the form
![\[
(1+x)^\alpha\ \square \ 1 + \alpha x + \binom\alpha2 x^2 + \cdots + \binom\alpha k x^k.\qquad(2)
\]](//latex.artofproblemsolving.com/f/5/b/f5b0f5ee3edf4d88e6614feafb849772ea55b942.png)
(Here 
, which allows for to be a non-integer.) Assume . Then the following hold.
, 
holds iff 
.
Proof. The reference [1] uses a proof by induction, but we'll instead go for a proof via Taylor's Theorem with Remainder. For ease of typesetting, let
be the right hand side of 
. Recall that is the 
order Taylor polynomial for the function 
. It follows that
for some real number 
between 
and 
. By assumption, 
, so 
is nonnegative. That is,
This proves part 1 of the original theorem, and parts 2 and 3 follow by analogous reasoning. 
References:
[1] Gerber, L. An Extension of Bernoulli's inequality. American Mathematical Monthly 75, 875 - 876 (1968).
Recall the statement of Bernoulli's Inequality:
whenever 
and 
(with the reverse inequality occuring when 
). In certain special cases, this is easy to see; for example, when 
and 
,![\[
(1+x)^3 = 1 + 3x + 3x^2 + x^3 \geq 1 + 3x.
\]](http://latex.artofproblemsolving.com/2/e/0/2e02f384a742ae503384c9914978968032b61ff8.png)
However, such a naive process does not necessarily work when 
is not an integer. For example, consider the case 
, which has Taylor expansion![\[
\sqrt{1+x} = 1 + \frac x2 - \frac18x^2 + \frac{1}{16}x^3 - \frac{5}{128}x^4 + \frac{7}{256}x^5 - \cdots.
\]](http://latex.artofproblemsolving.com/7/a/e/7ae61d9f990ed9a0094bfbe7ed57c7c45ef91441.png)
Here the Taylor expansion consists of both positive and negative coefficients, so more subtle analysis is needed to make the above approach work. Nevertheless, Bernoulli's Inequality does say that 
for all 
.At this point, we raise a natural question: does a "quadratic" version of Bernoulli's Inequality
![\[
(1+x)^r \geq 1 + rx + \frac{r(r-1)}2x^2\qquad(1)
\]](http://latex.artofproblemsolving.com/0/b/a/0bab5730a1150914079f08711eeb1858f573488a.png)
hold in general? What about higher approximations? It turns out the general criterion is surprisingly simple.Theorem ([1]). Let
be an arbitrary real number, and consider a hypothetical inequality of the form![\[
(1+x)^\alpha\ \square \ 1 + \alpha x + \binom\alpha2 x^2 + \cdots + \binom\alpha k x^k.\qquad(2)
\]](http://latex.artofproblemsolving.com/f/5/b/f5b0f5ee3edf4d88e6614feafb849772ea55b942.png)
(Here 
, which allows for to be a non-integer.) Assume . Then the following hold.
- If the next omitted term
of is positive, the LHS is greater than the RHS; - If the next omitted term is zero, the LHS and RHS are equal;
- If the next omitted term is negative, the LHS is less than the RHS.
of , 
holds iff 
.Proof. The reference [1] uses a proof by induction, but we'll instead go for a proof via Taylor's Theorem with Remainder. For ease of typesetting, let
be the right hand side of 
. Recall that is the 
order Taylor polynomial for the function 
. It follows that
for some real number 
between 
and 
. By assumption, 
, so 
is nonnegative. That is,
This proves part 1 of the original theorem, and parts 2 and 3 follow by analogous reasoning. 
References:
[1] Gerber, L. An Extension of Bernoulli's inequality. American Mathematical Monthly 75, 875 - 876 (1968).
This post has been edited 5 times. Last edited by djmathman, Sep 16, 2024, 4:55 PM
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