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I was reading a proof where at a certain point the prover uses the following inequality

$$A + 1 \leq e^A$$

which in my opinion needs also a proof to be used around. I think I'm missing some important fundamental property which is well-known, but I'm not sure.

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    Consider the function $f(A)=e^A-A-1$ and find its minimum by differentiation. – Michael May 23 '16 at 21:18
  • This is clear geometrically: the function $e^x$ has tangent line $y = x+1$ at $x = 0$ and it lies everywhere above any tangent line (look at the graphs, or use the second derivative test), so $e^x \geq x+1$ for all $x$, in fact with equality only for $x = 0$. – KCd May 23 '16 at 21:54
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    Why are the answerers only doing the case of $A>0$? – GEdgar May 23 '16 at 21:59
  • @GEdgar, because that's what the OP asks for? (Note, my answer addresses the general case in a remark at the end.) – Barry Cipra May 23 '16 at 22:02
  • @BarryCipra ... OK, in the title, but not the text itself ... – GEdgar May 23 '16 at 22:03
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    @GEdgar, also, all the answers need a little extra reasoning to handle the case $A\lt0$. – Barry Cipra May 23 '16 at 22:05

4 Answers4

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In the simple case of $A>0$, it suffices to just look at the taylor series definition:

$$1+A\leq 1+A+A^2/2!+A^3/3!+\cdots=e^A$$

Alex R.
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Since $(e^x)' = e^x$ for all $x$, and $e^x > 1$ for $x > 0$, we have, for $x > 0$,

$\begin{array}\\ e^x-1 &=\int_0^x e^t dt\\ &> \int_0^x 1 dt \qquad\text{since } e^x > 1 \text{ for } x > 0\\ &=x\\ \end{array} $

marty cohen
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For $n\in\mathbb{N}$ and $x\gt-1$, we have

$$(1+x)^n\ge1+nx$$

which is easily proved by induction on $n$:

$$\begin{align} (1+x)^n\ge1+nx\implies(1+x)^{n+1}&=(1+x)(1+x)^n\\ &\ge(1+x)(1+nx)\\ &=1+(n+1)x+nx^2\\ &\ge1+(n+1)x \end{align}$$

It follows that, if $A\gt0$, then

$$\left(1+{A\over n}\right)^n\ge1+n\left(A\over n\right)=1+A$$

and thus

$$e^A=\lim_{n\to\infty}\left(1+{A\over n}\right)^n\ge\lim_{n\to\infty}(1+A)=1+A$$

Remark: The inequality actually holds for all $A$, not just $A\gt0$, by the same proof, with the only extra wrinkle being that one needs to note that $A/n\gt-1$ for all $n\gt|A|$.

Barry Cipra
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$A+1=e^A$ at $A=0$.

$\frac{d(A+1)}{dA}=1$, while $\frac{d(e^A)}{dA}=e^A$.

$e^A>1 \forall A>0$.

So they start off equivalent, but the RHS grows faster at all points than the LHS.

Neil
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