Real Analysis Proof: (1+x)^y ≤ 1+ x^y for 0<y≤1 - Homework Help

[bobcat817]

Homework Statement

Let y be a fixed real number satisfying 0<y\leq1. Prove that (1+x)^{y}\leq1+ x^{y} for all x\geq0.

Homework Equations

I'm not sure.

The Attempt at a Solution

The hint given with the problem states that the derivative of x^{y} is yx^{y-1}. My first thought is that I'm supposed to show that they are both strictly increasing, but I don't really know what that would help me with.

I'm not really looking for an answer so much as a bit of direction.

 

[Pere Callahan]

Homework Statement

Let y be a fixed real number satisfying 0<y\leq1. Prove that (1+x)^{y}\leq1+ x^{y} for all x\geq0.
.

Are you sure you didn't reverse the inequality sign?
Take f(x)=(1+x)^y-1-x^y and show that f' is positive for all positive x. This implies that f(x) is positive for all x because f(0)=0.

 

[bobcat817]

Oh yes. I am quite certain that the inequality is right. Doing a few test cases shows that it is the correct inequality. So, basically, I end up with:

-1<p\leq0 Then letting p=|p|

and f'(x)=y\frac{1}{x^{p}}-\frac{1}{(1+x)^{p}}. And since x\geq0 for all x, and since x<x+1 for all x, x^{p}\leq(x+1)^{p} and so \frac{1}{(x+1)^{p}}\leq\frac{1}{x^{p}} giving that f'(x) is always positive and with f(0)=0, the intended result.

I trust that this is in the right direction, and thank you very much for your response.

 

[bobcat817]

I suppose I should do a separate case for p=0.