Qi) Prove Bernoulli's inequality If $h> -1$,

then $ (1+h)^n \geq 1+nh$

Qii) why is this Trivial is $h>0 $

Something i have always been lucky with is having a lot of intuition to go on with most theorems and lemmas that i see stated but this one is a bit different. For one looking at the Equality sign makes my head hurt i would like to hope that there is only equality when $h=0$ or when $n=1$ or when n=2 and $h = - \dfrac {1} {2}$ but it really makes my head hurt thinking about greater values of n where h is negative. i feel like there should be one value h that makes it true for each $ n \in \mathbb{N} $ Can anyone tell if i am missing values for n and h that result in an equality occurring? Perhaps it wouldn't even be all that difficult to prove there existence using contraction and induction; it may even be possible to define them using the binomial theorem.

As for Qii) This is fairly obvious expanding the left side we get crap $+nh +1 $ but unlike when h<0 the sum of crap $> 0$ so subtracting 1+nh from both side we get crap $>0 \geq 0$ which is obviously true.

Im not a big fan of approaching things inductively unless i absolutely have to especially on an inequality.

Although i wouldn't mind someone pointing out why my induction proof went a little wrong; i only made an attempt at proving it inductively below so you guys won't think i am lazy/ so someone won't write down a proof by induction.
If its possible could someone show me how to prove Qi) without using induction.

Proving Qi) inductively is fairly easy base case $n=1$ we have $ (1+h) \geq (1+h) $ thus base case is true assume true for all k want to show true for all $ (k+1) $

Now we want to show that $ (1+h) (1+h)^k \geq 1+(k+1)h$ is true.

This reduces to $ (1+h) (1+h)^k \geq (1+hk) +h$ using the k case we can reduce this to be true whenever the inequality $ (1+h) \geq +h$ which is always true for all h this implies that the inequality holds for all $n\in \mathbb{N} $ and $ \forall h \in \mathbb{R} $ but that's actually not true and would only be true if n were even since n can be odd the left side can be bigger in magnitude but actually be negative if $(1+h) <0$ luckily for me $ h> -1 $ is a condition.

  • 1
    $\begingroup$ Let $x = 1 + h$, so you want to show $x^n \geq 1 + n(x-1)$ for $x > 0$. The graph of $y = x^n$ along the positive $x$-axis is concave up, so it lies above any of its tangent lines over this interval except at the point of tangency. Figure out the equation of the tangent line to $y = x^n$ at the point $(1,1)$. $\endgroup$ – KCd Jan 2 '15 at 18:42
  • $\begingroup$ Do you mean like a Taylor series expansion ( though i guess you would of set it up as 1+h = x-1 if that's what you wanted me to do.) or just the normal derivative? $\endgroup$ – Faust Jan 2 '15 at 19:17
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    $\begingroup$ Well, in a sense I guess the answer is "yes" because the first two terms in the Taylor expansion area the expression for the tangent line, but you don't need to think about Taylor expansions. Just figure out the slope of the tangent line at the point $(1,1)$ and then the equation of the line through $(1,1)$ with that slope. Say it is $y = mx+b$. Since the graph of $y = x^n$ lies above any of its tangent lines over the interval $(0,\infty)$ except at $(1,1)$ itself, we get $x^n \geq mx + b$ for $x > 0$, with equality only at $x = 1$ (corresponding to $h = 0$ in the original notation). $\endgroup$ – KCd Jan 2 '15 at 20:09
  • $\begingroup$ Thats very clever I really like that approach thanks! $\endgroup$ – Faust Jan 2 '15 at 21:04

I'll assume you mean $n$ is an integer.

Here's how one can easily go about a proof by induction. The proof for $n=1$ is obvious.

Assume the case is established for $n$ then, $(1+h)^{n+1}=(1+h)^n(1+h)\geq(1+nh)(1+h)=1+(n+1)h+nh^2\geq 1+(n+1)h$

The part where $h\geq-1$ is in using the $\geq$ where we assume that $(1+h)$ is positive.

There is a way I know to prove it without induction that extends to real numbers instead of only integers.

Let $f(x)=x^\alpha-\alpha x+\alpha-1$.


$f'(x)=0$ at $x=1$ and it is obvious that $f(x)$ has a local minimum when $\alpha<0$ or $1<\alpha$ and $x>0$

At $x=1$, $f(x)=0$ so $f(x)\geq0$

Set $x=(1+h)$

Now, you have:

$(1+h)^\alpha-\alpha-\alpha h+\alpha-1=(1+h)^\alpha-\alpha h-1\geq 0$ and we can deduce Bernoulli's Inequality as a special case.

  • $\begingroup$ Very Clever i realized how to fix my proof w.o using the change you did on the inductive end. i really like the second proof w.o induction though thanks so much! $\endgroup$ – Faust Jan 2 '15 at 19:40

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