An illustration of Bernoulli's inequality, with the graphs of
y = ( 1 + x ) r {\displaystyle y=(1+x)^{r}} and
y = 1 + r x {\displaystyle y=1+rx} shown in red and blue respectively. Here,
r = 3. {\displaystyle r=3.} In real analysis , Bernoulli's inequality (named after Jacob Bernoulli ) is an inequality that approximates exponentiations of 1 + x .
The inequality states that
( 1 + x ) r ≥ 1 + r x {\displaystyle (1+x)^{r}\geq 1+rx} for every integer r ≥ 0 and every real number x ≥ −1 [1] r is even , then the inequality is valid for all real numbers x . The strict version of the inequality reads
( 1 + x ) r > 1 + r x {\displaystyle (1+x)^{r}>1+rx} for every integer r ≥ 2 and every real number x ≥ −1 with x ≠ 0.
There is also a generalized version that says for every real number r ≥ 1 and real number x ≥ -1,
( 1 + x ) r ≥ 1 + r x , {\displaystyle (1+x)^{r}\geq 1+rx,} while for 0 ≤ r ≤ 1 and real number x ≥ -1,
( 1 + x ) r ≤ 1 + r x . {\displaystyle (1+x)^{r}\leq 1+rx.} Bernoulli's inequality is often used as the crucial step in the proof of other inequalities. It can itself be proved using mathematical induction , as shown below.
History [ edit ] Jacob Bernoulli first published the inequality in his treatise "Positiones Arithmeticae de Seriebus Infinitis" (Basel, 1689), where he used the inequality often.[2]
According to Joseph E. Hofmann, Über die Exercitatio Geometrica des M. A. Ricci (1963), p. 177, the inequality is actually due to Sluse in his Mesolabum (1668 edition), Chapter IV "De maximis & minimis".[2]
Proof of the inequality [ edit ] We proceed with mathematical induction in the following form:
we prove the inequality for r ∈ { 0 , 1 } {\displaystyle r\in \{0,1\}} from validity for some r we deduce validity for r+2 . For r = 0,
( 1 + x ) 0 ≥ 1 + 0 x {\displaystyle (1+x)^{0}\geq 1+0x} is equivalent to 1 ≥ 1 which is true as required.
Similarly, for r = 1 we have
( 1 + x ) r = 1 + x ≥ 1 + x = 1 + r x . {\displaystyle (1+x)^{r}=1+x\geq 1+x=1+rx.} Now suppose the statement is true for r = k :
( 1 + x ) k ≥ 1 + k x . {\displaystyle (1+x)^{k}\geq 1+kx.} Then it follows that
( 1 + x ) k + 2 = ( 1 + x ) k ( 1 + x ) 2 ≥ ( 1 + k x ) ( 1 + 2 x + x 2 ) by hypothesis and ( 1 + x ) 2 ≥ 0 = 1 + 2 x + x 2 + k x + 2 k x 2 + k x 3 = 1 + ( k + 2 ) x + k x 2 ( x + 2 ) + x 2 ≥ 1 + ( k + 2 ) x {\displaystyle {\begin{aligned}(1+x)^{k+2}&=(1+x)^{k}(1+x)^{2}\\&\geq (1+kx)\left(1+2x+x^{2}\right)\qquad \qquad \qquad {\text{ by hypothesis and }}(1+x)^{2}\geq 0\\&=1+2x+x^{2}+kx+2kx^{2}+kx^{3}\\&=1+(k+2)x+kx^{2}(x+2)+x^{2}\\&\geq 1+(k+2)x\end{aligned}}} since x 2 ≥ 0 {\displaystyle x^{2}\geq 0} x + 2 ≥ 0 {\displaystyle x+2\geq 0} r .
Generalization [ edit ] The exponent r can be generalized to an arbitrary real number as follows: if x > −1, then
( 1 + x ) r ≥ 1 + r x {\displaystyle (1+x)^{r}\geq 1+rx} for r ≤ 0 or r ≥ 1, and
( 1 + x ) r ≤ 1 + r x {\displaystyle (1+x)^{r}\leq 1+rx} for 0 ≤ r ≤ 1.
This generalization can be proved by comparing derivatives . Again, the strict versions of these inequalities require x ≠ 0 and r ≠ 0, 1.
Related inequalities [ edit ] The following inequality estimates the r -th power of 1 + x from the other side. For any real numbers x , r with r > 0, one has
( 1 + x ) r ≤ e r x , {\displaystyle (1+x)^{r}\leq e^{rx},} where e = 2.718... . This may be proved using the inequality (1 + 1/k )k e .
Alternative form [ edit ] An alternative form of Bernoulli's inequality for t ≥ 1 {\displaystyle t\geq 1} 0 ≤ x ≤ 1 {\displaystyle 0\leq x\leq 1}
( 1 − x ) t ≥ 1 − x t . {\displaystyle (1-x)^{t}\geq 1-xt.} This can be proved (for integer t) by using the formula for geometric series : (using y=1-x)
t = 1 + 1 + ⋯ + 1 ≥ 1 + y + y 2 + … + y t − 1 = 1 − y t 1 − y {\displaystyle t=1+1+\dots +1\geq 1+y+y^{2}+\ldots +y^{t-1}={\frac {1-y^{t}}{1-y}}} or equivalently x t ≥ 1 − ( 1 − x ) t . {\displaystyle xt\geq 1-(1-x)^{t}.}
Alternative Proof [ edit ] Using AM-GM
An elementary proof for 0 ≤ r ≤ 1 {\displaystyle 0\leq r\leq 1} Weighted AM-GM .
Let λ 1 , λ 2 {\displaystyle \lambda _{1},\lambda _{2}} 1 , 1 + x {\displaystyle 1,1+x} λ 1 , λ 2 {\displaystyle \lambda _{1},\lambda _{2}}
λ 1 ⋅ 1 + λ 2 ⋅ ( 1 + x ) λ 1 + λ 2 ≥ ( 1 + x ) λ 2 λ 1 + λ 2 {\displaystyle {\dfrac {\lambda _{1}\cdot 1+\lambda _{2}\cdot (1+x)}{\lambda _{1}+\lambda _{2}}}\geq {\sqrt[{\lambda _{1}+\lambda _{2}}]{(1+x)^{\lambda _{2}}}}}
Note that
λ 1 ⋅ 1 + λ 2 ⋅ ( 1 + x ) λ 1 + λ 2 = λ 1 + λ 2 + λ 2 x λ 1 + λ 2 = 1 + λ 2 λ 1 + λ 2 x {\displaystyle {\dfrac {\lambda _{1}\cdot 1+\lambda _{2}\cdot (1+x)}{\lambda _{1}+\lambda _{2}}}={\dfrac {\lambda _{1}+\lambda _{2}+\lambda _{2}x}{\lambda _{1}+\lambda _{2}}}=1+{\dfrac {\lambda _{2}}{\lambda _{1}+\lambda _{2}}}x}
and
( 1 + x ) λ 2 λ 1 + λ 2 = ( 1 + x ) λ 2 λ 1 + λ 2 {\displaystyle {\sqrt[{\lambda _{1}+\lambda _{2}}]{(1+x)^{\lambda _{2}}}}=(1+x)^{\frac {\lambda _{2}}{\lambda _{1}+\lambda _{2}}}}
so our inequality is equivalent to
1 + λ 2 λ 1 + λ 2 x ≥ ( 1 + x ) λ 2 λ 1 + λ 2 {\displaystyle 1+{\dfrac {\lambda _{2}}{\lambda _{1}+\lambda _{2}}}x\geq (1+x)^{\frac {\lambda _{2}}{\lambda _{1}+\lambda _{2}}}}
After substituting r = λ 2 λ 1 + λ 2 {\displaystyle r={\dfrac {\lambda _{2}}{\lambda _{1}+\lambda _{2}}}} 0 ≤ r ≤ 1 {\displaystyle 0\leq r\leq 1}
1 + r x ≥ ( 1 + x ) r {\displaystyle 1+rx\geq (1+x)^{r}}
Using the formula for geometric series
Bernoulli's inequality
( 1 + x ) r ≥ 1 + r x {\displaystyle (1+x)^{r}\geq 1+rx} (1)
is equivalent to
( 1 + x ) r − 1 − r x ≥ 0 {\displaystyle (1+x)^{r}-1-rx\geq 0} (2)
and by the formula for geometric series (using y=1+x) we get
( 1 + x ) r − 1 = y r − 1 = ( ∑ k = 0 r − 1 y k ) ⋅ ( y − 1 ) = ( ∑ k = 0 r − 1 ( 1 + x ) k ) ⋅ x {\displaystyle (1+x)^{r}-1=y^{r}-1=\left(\sum _{k=0}^{r-1}y^{k}\right)\cdot (y-1)=\left(\sum _{k=0}^{r-1}(1+x)^{k}\right)\cdot x} (3)
which leads to
( 1 + x ) r − 1 − r x = ( ( ∑ k = 0 r − 1 ( 1 + x ) k ) − r ) ⋅ x = ( ∑ k = 0 r − 1 ( ( 1 + x ) k − 1 ) ) ⋅ x ≥ 0 {\displaystyle (1+x)^{r}-1-rx=\left(\left(\sum _{k=0}^{r-1}(1+x)^{k}\right)-r\right)\cdot x=\left(\sum _{k=0}^{r-1}\left((1+x)^{k}-1\right)\right)\cdot x\geq 0} (4 )
Now if x ≥ 0 {\displaystyle x\geq 0} ( 1 + x ) k − 1 ≥ 0 {\displaystyle (1+x)^{k}-1\geq 0} 0 {\displaystyle 0} LHS of (4
If 0 ≥ x ≥ − 2 {\displaystyle 0\geq x\geq -2} 1 ≥ ( 1 + x ) k {\displaystyle 1\geq (1+x)^{k}} ( 1 + x ) k − 1 {\displaystyle (1+x)^{k}-1} 4
Using Binomial theorem
(1) For x > 0 ( 1 + x ) r = 1 + r x + ( r 2 ) x 2 + . . . + ( r r ) x r {\displaystyle (1+x)^{r}=1+rx+{\tbinom {r}{2}}x^{2}+...+{\tbinom {r}{r}}x^{r}} ( r 2 ) x 2 + . . . + ( r r ) x r ≥ 0 {\displaystyle {\tbinom {r}{2}}x^{2}+...+{\tbinom {r}{r}}x^{r}\geq 0}
Thus, ( 1 + x ) r ≥ 1 + r x {\displaystyle (1+x)^{r}\geq 1+rx}
(2) For x = 0 ( 1 + x ) r = 1 + r x {\displaystyle (1+x)^{r}=1+rx}
(3) For −1 ≤ x < 0 , let y = −x 0 < y ≤ 1
Replace x −y ( 1 − y ) r = 1 − r y + ( r 2 ) y 2 + . . . + ( − 1 ) r ( r r ) y r {\displaystyle (1-y)^{r}=1-ry+{\tbinom {r}{2}}y^{2}+...+(-1)^{r}{\tbinom {r}{r}}y^{r}}
Also, according to the binomial theorem, 0 = 1 − r + ( r 2 ) + . . . + ( − 1 ) r ( r r ) {\displaystyle 0=1-r+{\tbinom {r}{2}}+...+(-1)^{r}{\tbinom {r}{r}}}
then ( r 2 ) + . . . + ( − 1 ) r ( r r ) = r − 1 ≥ 0 {\displaystyle {\tbinom {r}{2}}+...+(-1)^{r}{\tbinom {r}{r}}=r-1\geq 0}
Notice that y 2 ≥ y 3 ≥ . . . ≥ y r {\displaystyle y^{2}\geq y^{3}\geq ...\geq y^{r}}
Therefore, we can see that each binomial term ( r n ) {\displaystyle {\tbinom {r}{n}}} y n {\displaystyle y^{n}}
For that reason, ( r 2 ) y 2 + . . . + ( − 1 ) r ( r r ) y r ≥ 0 {\displaystyle {\tbinom {r}{2}}y^{2}+...+(-1)^{r}{\tbinom {r}{r}}y^{r}\geq 0}
Hence, ( 1 − y ) r = 1 − r y + ( r 2 ) y 2 + . . . + ( − 1 ) r ( r r ) y r ≥ 1 − r y {\displaystyle (1-y)^{r}=1-ry+{\tbinom {r}{2}}y^{2}+...+(-1)^{r}{\tbinom {r}{r}}y^{r}\geq 1-ry}
Replace y −x ( 1 + x ) r ≥ 1 + r x {\displaystyle (1+x)^{r}\geq 1+rx}
Notice that by using binomial theorem, we can only prove the cases when r is a positive integer or zero.
References [ edit ] Carothers, N.L. (2000). Real analysis 9 . ISBN 978-0-521-49756-5 Bullen, P. S. (2003). Handbook of means and their inequalities . Dordercht [u.a.]: Kluwer Academic Publ. p. 4. ISBN 978-1-4020-1522-9 Zaidman, S. (1997). Advanced calculus : an introduction to mathematical analysis . River Edge, NJ: World Scientific. p. 32. ISBN 978-981-02-2704-3 External links [ edit ] 
![\dfrac{\lambda_1\cdot 1+\lambda_2\cdot (1+x)}{\lambda_1+\lambda_2}\ge \sqrt[\lambda_1+\lambda_2]{(1+x)^{\lambda_2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3412a548d2f2339a456161a853a638b38904bf2c)
![\sqrt[\lambda_1+\lambda_2]{(1+x)^{\lambda_2}}=(1+x)^{\frac{\lambda_2}{\lambda_1+\lambda_2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eaa259e4f097b8365b8a3f585ec1f3f87f07b7fe)
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