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.
Answer
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 $f(x)=x^\alpha-\alpha x+\alpha-1$.
$f'(x)=\alpha*(x^{\alpha-1}-1)$
$f'(x)=0$ 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.
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