- #1

MikeyA

- 4

- 0

## Homework Statement

The question reads very simply, show that e

^{x}[tex]\geq[/tex]1+x [tex]\forall[/tex] x > 0

## Homework Equations

None to speak of. I am not allowed to use calculus, and this is why I am having problems.

## The Attempt at a Solution

I tried to break it up into cases:

When x=0, e

^{x}=e

^{0}=1 1+x=1+0=1

Hence e

^{x}=1+x.

Then I decided to take the natural log of both sides and try my hand at the other three cases (which I think should be 0<x<1, x=1, and x>1):

ln(1+x)[tex]\leq[/tex]lne

^{x}

<=> ln(1+x)[tex]\leq[/tex]x

ln2<x is easy, but I cannot even imagine how I would be able to solve the other cases.

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