- #1
r0bHadz
- 194
- 17
Homework Statement
Prove bernullis inequality: If h>-1 then (1+h)^n ≥ 1+ nh
Homework Equations
The Attempt at a Solution
How can I prove something that is false for h =1 n<1 ?
r0bHadz said:(1+h)^n ≥ h^n +1
because (1+h)^n will be at least h^n + xh^(n-1)+...+1
I believe h^n + 1 ≥ nh+1
all I would have to do is show x^y ≥ xy and the proof would be done. Do you think my method is sound?
Delta2 said:The usual proof of this inequality is by induction on n. The proof by induction works even if ##h>-1\Rightarrow h+1>0##.(while your proof at post #3 works only if ##h>1##.)
This was the first proof by induction that I was taught , I remembered my high school years now.
Ray Vickson said:Right: induction is great at an elementary level. A different (calculus-based) method shows the result to be true if ##h >-1## and ##n \geq 1## is real. However, I guess this would not be appropriate in a "pre-calculus" forum.
Ray Vickson said:Right: induction is great at an elementary level. A different (calculus-based) method shows the result to be true if ##h >-1## and ##n \geq 1## is real. However, I guess this would not be appropriate in a "pre-calculus" forum.
I think the logarithm plus power series could work, but there has to be a more elegant solution with the mean value theorem or so.Delta2 said:I am interested in your calculus based method that proves it for ##n## real (and ##h>-1##). Can you give any hint?
This proof has been brought to you by Hölder & Young.StoneTemplePython said:...
##(1 +ph)^\frac{1}{p}##
## = (1 +ph)^\frac{1}{p}\cdot 1##
## = (1 +ph)^\frac{1}{p}\cdot 1^\frac{p-1}{p}##
## \leq \frac{1}{p} (1 +ph) + \frac{p-1}{p}(1)##
## = 1 +h##
by ##\text{GM} \leq \text{AM}##
(you just need analytic tools to prove the strong form of this geometric vs arithmetic mean inequality)
fresh_42 said:
StoneTemplePython said:I'll take that at face value and show my preferred approach for the general case (a lot nicer than wikipedia in my view):
for real ##p \geq 1## and ##h \in(-1,\infty)## we have
##1 +ph \leq (1 +h)^p##
the right side is always positive so we only need to put effort into proving the case where the left hand side is positive.
in which case we can take pth roots and seek to prove the equivalent ##(1 +ph)^\frac{1}{p} \leq 1 +h##.
So we have
##(1 +ph)^\frac{1}{p}##
## = (1 +ph)^\frac{1}{p}\cdot 1##
## = (1 +ph)^\frac{1}{p}\cdot 1^\frac{p-1}{p}##
## \leq \frac{1}{p} (1 +ph) + \frac{p-1}{p}(1)##
## = 1 +h##
by ##\text{GM} \leq \text{AM}##
(you just need analytic tools to prove the strong form of this geometric vs arithmetic mean inequality)
Bernoulli's inequality is a mathematical inequality that states that for any real number x greater than or equal to -1 and any positive integer n, (1+x)^n ≥ 1+nx.
Bernoulli's inequality was discovered by Swiss mathematician Jacob Bernoulli in the late 17th century.
Bernoulli's inequality is an important tool in mathematical analysis and is used in many areas of mathematics, including calculus, number theory, and statistics. It also has applications in physics and engineering.
Bernoulli's inequality can be proved using mathematical induction. The base case (n=1) is trivial, and then the inductive step involves showing that if the inequality holds for some integer k, then it also holds for k+1. This completes the proof by induction.
Bernoulli's inequality can be applied to compound interest calculations, where the interest rate is greater than 0. It can also be used to prove the AM-GM inequality, which is used in statistics to find the arithmetic mean of a set of numbers. Additionally, Bernoulli's inequality is used in the proof of the binomial theorem, which has applications in probability and combinatorics.