Can Bernoulli's Inequality Be Proven for All Real Numbers?

[r0bHadz]

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 ?

 

[fresh_42]

$n$ is a natural number. So even if you consider $n=0$ to be a natural number, then it is the only one smaller than one. But $(1+h)^0=1 \geq 1+0 \cdot h$ is true. Prove it for $n=1,2,3,\ldots$

 

[r0bHadz]

(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?

 

[Ray Vickson]

(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?

I think you are saying that
$$(1+h)^n = 1 + nh +\frac{1}{2} n(n-1) h^2 + \frac{1}{3!} n(n-1)(n-2) h^3 + \cdots + h^n \geq 1 + nh$$ Certainly, this argument is obviously true when $h \geq 0$ (because we just drop a number of non-negative terms to get to $1+nh$). However, the reasoning breaks down when $-1 < h < 0$.

The final result is still true in that case, but it needs a different type of argument.

 

[Delta2]

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 $h>2$ (to be more precise your method works for some p $1<p<2$ such that $h>p$.)
This was the first proof by induction that I was taught , I remembered my high school years now.

 

[Ray Vickson]

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.

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.

 

[StoneTemplePython]

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.

if we constrain $n\geq 1$ to be rational, then the result follows from basic $\text{GM}\leq \text{AM}$ (proven by one of many elementary means). However to consider irrational values, some kind of analytic tools end up being needed, naturally.

- - - -
edit:
for avoidance of doubt $\text{GM}\leq \text{AM}$ in its full power with real weights gives the proof here over real $n\geq 1$. The issue is that you need to use analytic tools (limits or better: convexity of exponential function) to prove this more general form of $\text{GM}\leq \text{AM}$ which I think is beyond the scope for OP right now.

 

[Delta2]

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 am interested in your calculus based method that proves it for $n$ real (and $h>-1$). Can you give any hint?

 

[fresh_42]

I am interested in your calculus based method that proves it for $n$ real (and $h>-1$). Can you give any hint?

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]

After I looked at Wikipedia's entry for Bernoulli's inequality, I think a way to prove it is to consider the function $f(h)=(1+h)^n-1-hn$ and prove that this function is increasing using derivatives, that is prove that $f'(h)\geq 0$. Then the result will follow from $h>-1\Rightarrow f(h)\geq f(-1)=n-1\geq 0$

EDIT: Turns out that this $f(h)$ is increasing for $h>0$ and is decreasing for $-1<h<0$ but because $f(0)=0$ the method still works.

 

[r0bHadz]

Sorry forgot to press "solved" guys. Doing this with induction on n took me 1 minute. I don't know why the author didn't specify that n is a natural number but I guess its assumed. I'll look into other ways of doing this problem as I advance inshaAllah

 

[StoneTemplePython]

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)

 

[fresh_42]

...
$(1 +ph)^\frac{1}{p}$
$= (1 +ph)^\frac{1}{p}\cdot 1$
$= (1 +ph)^\frac{1}{p}\cdot 1^\frac{p-1}{p}$
$\leq \frac{1}{p} (1 +ph) + \frac{p-1}{p}(1)$
$= 1 +h$

by $\text{GM} \leq \text{AM}$

(you just need analytic tools to prove the strong form of this geometric vs arithmetic mean inequality)

This proof has been brought to you by Hölder & Young.

(Sorry, I just need a mnemonic to remember this better and Hölder & Young sounded so funny in this context.)

 

[StoneTemplePython]

This proof has been brought to you by Hölder & Young.

(Sorry, I just need a mnemonic to remember this better and Hölder & Young sounded so funny in this context.)

it is true that I used $p$ in a nod to Hölder's Inequality and these are Hölder conjugates in the exponents. That said, one doesn't actually need to know Hölder's Inequality (or Young) to get this -- I think Cauchy would have gotten it by just playing around with $\text{GM} \leq \text{AM}$.

 

[Ray Vickson]

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)

My personal preference would be to note that for any real $n > 1$ the function $f(h) = (1+h)^n$ is strictly convex on the set $(-1,\infty)$, so $f(h) > f(0) + f'(0) h$ for any $h \neq 0, h > -1.$ This requires a little bit of fairly elementary material on convexity; in this case we need to know that for a convex function $f(h),$ the tangent line to the graph of $y = f(h)$ lies on or below the graph at all legitimate $h \neq 0$, and lies strictly below if $f$ is strictly convex.