Proof by induction series help

[talolard]

Hey everyone, I have a question on some proof from an old exam. Please excuse my Latex, I have not yet mastered this.
Thanks for the help.
Tal

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Homework Statement

for any $$1>t>-1 and p \geq 1$$ prove that $$(1+t)^p + (1-t)^p \geq 2$$

Thats all the info the question has. I am assuming p is natural.

Homework Equations

The Attempt at a Solution

Proof by induction. it is easy to see that p=1 fullfills what is required. So we shall assum for n and show that this invokes n+1.
$$(1+t)^{n+1} + (1-t)^{n+1} =\sum^{i=0}_{n+1}(\stackrel{n+1}{i})t^i + \sum^{i=0}_{n+1}(\stackrel{n+1}{i}) t^i$$
$$= \sum^{i=0}_{n+1}(\stackrel{n+1}{i})t^i + \sum^{i=0}_{(n+1)/2} (\stackrel{n+1}{(2i)}) t^{2i} - \sum^{i=1}_{(n+1)/2} (\stackrel{n+1}{(2i +1)}) t^{2i+1} = 2 + \sum^{i=1}_{n+1}(\stackrel{n+1}{i}) t^{i} + \sum^{i=1}_{(n+1)/2} (\stackrel{n+1}{(2i)}) t^{2i} - \sum^{i=1}_{(n+1)/2} (\stackrel{n+1}{(2i+1)}) t^{2i+1}$$

Since 1>t>-1 its succesive powers are smaller then the previous ones. Then we can see that the two partial sums of even and odd powers, the even powers are always larger then the odd powers and there sum is positive. The full sum is also positive for any t>0. but if we look at the original problem then we can see that if t < 1 the equation remains the same and so we always have a sum of 2 + something positive.

I wonder if this is correct.

 

[JSuarez]

You don't have to assume that p is natural; in fact, this inequelity is much more general. Just observe that:

(1) $$(1+t)^p + (1-t)^p$$ is an even function, so you just have to worry with $$0\leq t \leq 1$$.

(2) What is the derivative of the above expression? What is its sign for $$0\leq t \leq 1$$? What does this tell you about the minimum value?

 

[talolard]

Wow.
That is so much simpler. Thanks.
I'd still be glad if you could tell me if my orignal way is correct.
Thanks
Tal

 

[JSuarez]

That expression is too messy to tell.