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Is this valid, and how can i prove that it holds.

Proof of $$e^x - 1 \geq x \text{ for } {x:-1 \leq x < 0}$$

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  • $\begingroup$ If you graph $y=e^x-1$ and $y=x$ together, it's pretty easy to see that the inequality holds not just for $-1\leq x<0$, but for all $x$. However, this isn't a proof. You can convert it into a proof by using the mean value theorem. It helps to look at the function $f(x)=e^x-x-1$, and ask what it would mean if we had $f(x)<0$ for some $x<0$ (again, using the MVT, and asking what we know about $f'(x)=e^x-1$). $\endgroup$ – Michael Weiss May 7 '14 at 14:07
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By taking the second derivative we see that the exponential function is convex on $\Bbb R$ hence its curve is above the line tangent at the point $(0,1)$ with equation $$y=x+1.$$ Hence we deduce that $$e^x\ge x+1,\qquad \forall x\in \Bbb R$$

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Let $f(x)=e^x-x-1$. Then $f'(x)=e^x-1\le 0\ \forall x\in (-\infty,0]\Rightarrow f(x)\ge f(0)=0\Rightarrow e^x-1\ge x\ \forall x\in [-1,0]$.

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Consider $f(x)=e^x-1-x$

$\Rightarrow f'(x)=e^x-1$

Now $x<0\Rightarrow e^x<1\Rightarrow e^x-1<0$

So $f'(x)<0\forall x\in [-1,0)$

Thus $f$ is decreasing in $[-1,0)$ and also $f(0)=0$

So $\forall x\in [-1,0)$ We have $x<0$ and so $f(x)\ge f(0)$ as $f$ decreasing

Thus $e^x-1-x\ge0\forall x\in [-1,0)$

$\Rightarrow e^x-1\ge x\forall x\in [-1,0)$

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Alternatively via power series, $e^x$ is given by power series

$e^x = \sum_{k=0}^\infty \frac{x^k}{k!}$

so

$e^x -x -1 = \sum_{k=2}^\infty \frac{x^k}{k!}$.

Note that $|x|^n \ge 0$ gives

$\sum_{k=2}^\infty \frac{x^k}{k!} = \sum_{k=2}^\infty \frac{(-1)^k |x|^k}{k!} \ge 0 $ for all x.

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