# Real Analysis Proof

## Homework Statement

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

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.

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## Homework Statement

Let y be a fixed real number satisfying 0<y$$\leq$$1. Prove that (1+x)$$^{y}$$$$\leq$$1+ x$$^{y}$$ for all x$$\geq$$0.
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Are you sure you didn't reverse the inequality sign?
Take f(x)=(1+x)y-1-xy and show that f' is positive for all positive x. This implies that f(x) is positive for all x because f(0)=0.

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$$\leq$$0 Then letting p=|p|

and f'(x)=y$$\frac{1}{x^{p}}$$-$$\frac{1}{(1+x)^{p}}$$. And since x$$\geq$$0 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.

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