Editing Bernoulli's inequality

{{Short description|Inequality about exponentiations of ''1+x''}}
[[File:Bernoulli inequality.svg|right|thumb|An illustration of Bernoulli's inequality, with the [[graph of a function|graph]]s of <math>y = (1+x)^r</math> and <math>y = 1+rx</math> shown in red and blue respectively. Here, <math>r=3.</math>]]
In [[mathematics]], '''Bernoulli's inequality''' (named after [[Jacob Bernoulli]]) is an [[inequality (mathematics)|inequality]] that approximates [[exponentiation]]s of <math>1+x</math>. It is often employed in [[real analysis]]. It has several useful variants:<ref>{{cite book|last=Brannan|first=D. A.|url=https://books.google.com/books?id=N8bL9lQUGJgC&pg=PA262|title=A First Course in Mathematical Analysis|publisher=Cambridge University Press|year=2006|isbn=9781139458955|page=20}}</ref>

== Integer exponent ==
* Case 1: <math>(1 + x)^r \geq 1 + rx</math> for every integer <math>r\geq 1</math> and real number <math>x\geq-1</math>. The inequality is strict if <math>x\neq 0</math> and <math>r\geq 2</math>.
* Case 2: <math>(1 + x)^r \geq 1 + rx</math> for every integer <math>r\geq 0</math> and every real number <math>x\geq -2</math>.<ref>Excluding the case {{math|1=''r'' = 0}} and {{math|1=''x'' = –1}}, or assuming that [[Zero to the power of zero|{{math|1=0<sup>0</sup> = 1}}]].</ref>
* Case 3: <math>(1 + x)^r \geq 1 + rx</math> for every [[parity (mathematics)|even]] integer <math>r\geq 0</math> and every real number <math>x</math>.

== Real exponent ==
* <math>(1 + x)^r \geq 1 + rx</math> for every real number <math>r\geq 1</math> and <math>x\geq -1</math>. The inequality is strict if <math>x\neq 0</math> and <math>r\neq 1</math>.
* <math>(1 + x)^r \leq 1 + rx</math> for every real number <math>0\leq r\leq 1</math> and <math>x\geq -1</math>.

==History==
Jacob Bernoulli first published the inequality in his treatise "Positiones Arithmeticae de Seriebus Infinitis" (Basel, 1689), where he used the inequality often.<ref name=autogenerated1>[http://hsm.stackexchange.com/a/1891 mathematics – First use of Bernoulli's inequality and its name – History of Science and Mathematics Stack Exchange<!-- Bot generated title -->]</ref>

According to Joseph E. Hofmann, Über die Exercitatio Geometrica des M. A. Ricci (1963), p.&nbsp;177, the inequality is actually due to Sluse in his Mesolabum (1668 edition), Chapter IV "De maximis & minimis".<ref name=autogenerated1 />

==Proof for integer exponent==

The  first case has a simple inductive proof:

Suppose the statement is true for <math>r=k</math>:

:<math>(1+x)^k \ge 1+kx. </math>

Then it follows that

:<math>
\begin{align}
(1+x)^{k+1} &= (1+x)^k(1+x) \\
&\ge (1+kx)(1+x) \\
&=1+kx+x+kx^2 \\ 
&=1+x(k+1)+kx^2 \\
&\ge 1+(k+1)x
\end{align}
</math>

Bernoulli's inequality can be proved for case 2, in which <math>r</math> is a non-negative integer and <math>x\ge-2</math>, using [[mathematical induction]] in the following form:
* we prove the inequality for <math>r\in\{0,1\}</math>,
* from validity for some ''r'' we deduce validity for <math>r+2</math>.

For <math>r=0</math>,

:<math>(1+x)^0 \ge 1+0x</math>

is equivalent to <math>1\geq 1</math> which is true.

Similarly, for <math>r=1</math> we have

:<math>(1+x)^r=1+x \ge 1+rx. </math>

Now suppose the statement is true for <math>r=k</math>:

:<math>(1+x)^k \ge 1+kx. </math>

Then it follows that

:<math>
\begin{align}
(1+x)^{k+2} &= (1+x)^k(1+x)^2 \\
&\ge (1+kx)\left(1+2x+x^2\right) \qquad\qquad\qquad\text{ by hypothesis and }(1+x)^2\ge 0 \\
&=1+2x+x^2+kx+2kx^2+kx^3 \\ 
&=1+(k+2)x+kx^2(x+2)+x^2 \\
&\ge 1+(k+2)x
\end{align}
</math>

since <math>x^2\ge 0</math> as well as <math>x+2\ge0</math>. By the modified induction we conclude the statement is true for every non-negative integer <math>r</math>.

By noting that if <math>x<-2</math>, then <math>1+rx</math> is negative gives case 3.

==Generalizations==

=== Generalization of exponent ===
The exponent <math>r</math> can be generalized to an arbitrary real number as follows: if <math>x>-1</math>, then

:<math>(1 + x)^r \geq 1 + rx</math>

for <math>r\leq 0</math> or <math>\geq 1</math>, and

:<math>(1 + x)^r \leq 1 + rx</math>

for <math>0\leq r\leq 1</math>.

This generalization can be proved by convexity (see below) or by comparing [[derivative]]s. The strict versions of these inequalities require <math>x\neq 0</math> and <math>r\neq 0, 1</math>.

The case <math>0 \leq r \leq 1 </math> can also be derived from the case <math>r\geq 1</math> by noting that (using the main case result) <math> \left(1 + \frac{x}{r}\right)^{r} \geq 1 + x = \left[(1+x)^{\frac{1}{r}}\right]^r </math> and by using the fact that <math>f(x) = x^r </math> is monotonic. We can conclude that <math>1 + x/r \geq (1+x)^{\frac{1}{r}}</math> for <math>r \geq 1</math>, therefore <math>(1 + x)^l \leq 1 + lx</math> for <math>0 < l = 1/r \leq 1</math>. The leftover case <math> l = 0 </math> is verified separately.

=== Generalization of base ===
Instead of <math>(1+x)^n</math> the inequality holds also in the form <math>(1+x_1)(1+x_2)\dots(1+x_r) \geq 1+x_1+x_2 + \dots + x_r</math> where <math>x_1, x_2, \dots , x_r</math> are real numbers, all greater than <math>-1</math>, all with the same sign. Bernoulli's inequality is a special case when <math>x_1 = x_2 = \dots = x_r = x</math>. This generalized inequality can be proved by mathematical induction.

{{collapse top| title=Proof}}
In the first step we take <math>n=1</math>. In this case the inequality <math>1+x_1 \geq 1 + x_1</math> is obviously true.

In the second step we assume validity of the inequality for <math>r</math> numbers and deduce validity for <math>r+1</math> numbers.

We assume that<math display="block">(1+x_1)(1+x_2)\dots(1+x_r) \geq 1+x_1+x_2 + \dots + x_r</math>is valid. After multiplying both sides with a positive number <math>(x_{r+1} + 1)</math> we get:

<math>\begin{alignat}{2}
(1+x_1)(1+x_2)\dots(1+x_r)(1+x_{r+1}) \geq & (1+x_1+x_2 + \dots + x_r)(1+x_{r+1})
\\
\geq & (1+x_1+x_2+ \dots + x_r) \cdot 1 + (1+x_1+x_2+ \dots + x_r) \cdot x_{r+1}
\\
\geq & (1+x_1+x_2+ \dots + x_r) + x_{r+1} + x_1 x_{r+1} + x_2 x_{r+1} + \dots + x_r x_{r+1}
\\ \end{alignat}</math>

As <math>x_1, x_2, \dots x_r, x_{r+1}</math> all have the same sign, the products <math>x_1 x_{r+1}, x_2 x_{r+1}, \dots x_r x_{r+1}</math> are all positive numbers. So the quantity on the right-hand side can be bounded as follows:<math display="block">(1+x_1+x_2+ \dots + x_r) + x_{r+1} + x_1 x_{r+1} + x_2 x_{r+1} + \dots + x_r x_{r+1} \geq 1+x_1+x_2+ \dots + x_r + x_{r+1},</math>what was to be shown.
{{cob}}

=== Strengthened version ===
The following theorem presents a strengthened version of the Bernoulli inequality, incorporating additional terms to refine the estimate under specific conditions. Let the expoent <math>r</math> be a nonnegative integer and let <math>x</math> be a real number with <math>x \ge -2</math>  if <math>r</math> is odd and greater than 1. Then

<math>(1 + x)^{r} \geq 1 + rx + \lfloor r/2 \rfloor x^2</math> 

with equality if and only if <math>r \in \{0, 1, 2\}</math> or <math>x \in \{-2, 0\}</math>.<ref>{{Cite journal |last=Bradley |first=David M. |date=2024-12-23 |title=A Stronger Version of Bernoulli’s Inequality |url=https://link.springer.com/article/10.1007/s00283-024-10396-5 |journal=The Mathematical Intelligencer |language=en |doi=10.1007/s00283-024-10396-5 |issn=0343-6993|url-access=subscription }}</ref> 

== Related inequalities ==
The following inequality estimates the <math>r</math>-th power of <math>1+x</math> from the other side. For any real numbers <math>x</math> and <math>r</math> with <math>r >0</math>, one has

:<math>(1 + x)^r \le e^{rx},</math>

where <math>e =</math> [[e (number)|2.718...]]. This may be proved using the inequality

:<math> \left(1 + \frac{1}{k}\right)^k < e.</math>

==Alternative form==
An alternative form of Bernoulli's inequality for  <math> t\geq 1 </math> and  <math> 0\le x\le 1 </math>  is:

:<math>(1-x)^t \ge 1-xt.</math>

This can be proved (for any integer <math>t</math>) by using the formula for [[geometric series]]: (using <math>y=1-x</math>)

:<math>t=1+1+\dots+1 \ge 1+y+y^2+\ldots+y^{t-1} = \frac{1-y^t}{1-y},</math>
or equivalently <math>xt \ge 1-(1-x)^t.</math>

==Alternative proofs==
===Arithmetic and geometric means===
An elementary proof for <math>0\le r\le 1</math> and <math>x \ge -1</math> can be given using [[Inequality of arithmetic and geometric means#Weighted AM–GM inequality|weighted AM-GM]].

Let <math>\lambda_1, \lambda_2</math> be two non-negative real constants. By weighted AM-GM on <math>1,1+x</math> with weights <math>\lambda_1, \lambda_2</math> respectively, we get

:<math>\dfrac{\lambda_1\cdot 1 + \lambda_2\cdot (1+x)}{\lambda_1+\lambda_2}\ge \sqrt[\lambda_1+\lambda_2]{(1+x)^{\lambda_2}}.</math>

Note that

:<math>\dfrac{\lambda_1\cdot 1 + \lambda_2\cdot (1+x)}{\lambda_1+\lambda_2}=\dfrac{\lambda_1+\lambda_2+\lambda_2x}{\lambda_1+\lambda_2}=1+\dfrac{\lambda_2}{\lambda_1+\lambda_2}x</math>

and

:<math>\sqrt[\lambda_1+\lambda_2]{(1+x)^{\lambda_2}} = (1+x)^{\frac{\lambda_2}{\lambda_1+\lambda_2}},</math>

so our inequality is equivalent to

:<math>1 + \dfrac{\lambda_2}{\lambda_1+\lambda_2}x \ge (1+x)^{\frac{\lambda_2}{\lambda_1+\lambda_2}}.</math>

After substituting <math>r = \dfrac{\lambda_2}{\lambda_1+\lambda_2}</math> (bearing in mind that this implies <math>0\le r\le 1</math>) our inequality turns into

:<math>1+rx \ge (1+x)^r</math>

which is Bernoulli's inequality for <math>0\le r\le 1</math>.

The case <math>r\ge 1</math> can be derived from <math>0\le r\le 1</math> in the same way as 
the case <math>0\le r\le 1</math> can be derived from <math>r\ge 1</math>, see above "Generalization of exponent".

===Geometric series===
Bernoulli's inequality

{{NumBlk|:|<math>(1+x)^r \ge 1+rx </math>|1}}

is equivalent to

{{NumBlk|:|<math>(1+x)^r - 1-rx \ge 0,</math>|2}}
and by the formula for [[geometric series]] (using ''y'' = 1&thinsp;+&thinsp;''x'') we get
{{NumBlk|:|<math> (1+x)^r - 1 = y^r-1 = \left(\sum^{r-1}_{k=0}y^k\right) \cdot (y-1) = \left(\sum^{r-1}_{k=0}(1+x)^k\right)\cdot x</math>|3}}  
which leads to
{{NumBlk|:|<math>(1+x)^r - 1-rx = \left(\left(\sum^{r-1}_{k=0}(1+x)^k\right) - r\right)\cdot x = \left(\sum^{r-1}_{k=0}\left((1+x)^k-1\right)\right)\cdot x \ge 0.</math>|{{EquationRef|4}}}}

Now if <math>x \ge 0</math> then by monotony of the powers each summand <math>(1+x)^k - 1 = (1+x)^k - 1^k \ge 0</math>, and therefore their sum is greater <math>0</math> and hence the product on the [[Sides of an equation|LHS]] of ({{EquationNote|4}}).

If <math> 0 \ge x\ge -2 </math> then by the same arguments <math>1\ge(1+x)^k</math> and thus
all addends <math>(1+x)^k-1</math> are non-positive and hence so is their sum. Since the product of two non-positive numbers is non-negative, we get again
({{EquationNote|4}}).

===Binomial theorem===
One can prove Bernoulli's inequality for ''x'' ≥ 0 using the [[binomial theorem]]. It is true trivially for ''r'' = 0, so suppose ''r'' is a positive integer. Then <math>(1+x)^r = 1 + rx + \tbinom r2 x^2 + ... + \tbinom rr x^r.</math> Clearly <math>\tbinom r2 x^2 + ... + \tbinom rr x^r \ge 0,</math> and hence <math>(1+x)^r \ge 1+rx</math> as required.

===Using convexity===
For <math>0\neq x> -1</math> the function <math>h(\alpha)=(1+x)^\alpha</math> is strictly convex. Therefore, for <math>0<\alpha<1</math> holds
<math>(1+x)^\alpha=h(\alpha)=h((1-\alpha)\cdot 0+\alpha\cdot 1)<(1-\alpha) h(0)+\alpha h(1)=1+\alpha x</math>
and the reversed inequality is valid for <math>\alpha<0</math> and <math>\alpha>1</math>.

Another way of using convexity is to re-cast the desired inequality to  <math>\log (1 + x) \geq \frac{1}{r}\log( 1 + rx)</math> for real <math>r\geq 1</math> and real  <math>x > -1/r</math>. This inequality can be proved using  the fact that the <math>\log</math>  function is concave, and  then using Jensen's inequality in the form <math> \log( p \, a + (1-p)b ) \geq p\log(a) + (1-p)\log(b) </math> to give:
<math>\log(1+x) = \log(\frac{1}{r}(1+rx)+\frac{r-1}{r}) \geq 
\frac{1}{r} \log (1+rx)+\frac{r-1}{r}\log 1 = \frac{1}{r} \log (1+rx)
</math> 
which is the desired inequality.

==Notes==
{{reflist}}

==References==
* {{cite book|last1=Carothers|first1=N.L.|title=Real analysis|url=https://archive.org/details/realanalysis0000caro/page/9|url-access=registration|year=2000|publisher=Cambridge University Press|location=Cambridge|isbn=978-0-521-49756-5|page=[https://archive.org/details/realanalysis0000caro/page/9 9]}}
* {{cite book|last1=Bullen|first1=P. S.|title=Handbook of means and their inequalities|url=https://archive.org/details/handbookmeansthe00bull|url-access=limited|year=2003|publisher=Kluwer Academic Publ.|location=Dordercht [u.a.]|isbn=978-1-4020-1522-9|page=[https://archive.org/details/handbookmeansthe00bull/page/n26 4]}}
* {{cite book|last1=Zaidman|first1=S.|title=Advanced calculus : an introduction to mathematical analysis|url=https://archive.org/details/advancedcalculus00zaid|url-access=limited|year=1997|publisher=World Scientific|location=River Edge, NJ|isbn=978-981-02-2704-3|page=[https://archive.org/details/advancedcalculus00zaid/page/n38 32]}}
* {{cite book|last1=[[Dragoslav Mitrinović|Mitrinović]]|first1=D. S.|title=Analytic Inequalities. In cooperation with [[Petar Vasić|P. M. Vasić]]|series = Die Grundlehren der mathematischen Wissenschaften| volume = 165| year=1970|publisher=[[Springer Science+Business Media|Springer Verlag]]|location=[[Berlin]], Heidelberg, New York |isbn=978-3-642-99972-7|doi=10.1007/978-3-642-99970-3|zbl=0199.38101 }}

== External links ==
* {{MathWorld |title= Bernoulli Inequality |urlname= BernoulliInequality}}
* [https://demonstrations.wolfram.com/BernoulliInequality/ Bernoulli Inequality] by Chris Boucher, [[Wolfram Demonstrations Project]].
* {{cite web|title=Introduction to Inequalities|url=https://www.mediafire.com/file/1mw1tkgozzu |author=Arthur Lohwater|year=1982|publisher=Online e-book in PDF format}}

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[[Category:Inequalities (mathematics)]]