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I need to prove (using the mean value theorem) that for all $x\in(0,\infty)$, the following inequality holds:

$\exp(-x)\geq 1 - x$


I don't know how the mean value theorem is applicable here, since we don't have a closed interval.

How do I prove the statement?

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    $\begingroup$ Make it your first step to make this into a closed interval. For instance, you could prove that $e^{-x} \geq 0$, and then show quickly that $0 \geq 1 - x$ for $x \in [1, \infty]$. That gives you a nice closed interval of $[0, 1]$ where you can do your proof. $\endgroup$ – Larry B. Jan 12 '17 at 20:07
  • $\begingroup$ A proof without MVT is at follows: $\exp(-x)$ is a convex function and the straight line $1-x$ is a tangent line at $0$. The graph of a differentaible convex function lies above (precisely not below) the secant. This argument works for all real $x$. We have even more: the inequality is strict, whenever $x\ne 0$. Indeedd, our function is in fact strictly convex. $\endgroup$ – szw1710 Jan 12 '17 at 20:29
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Hint: let $f(t)=e^{-t}$, and (for a fixed $x$) try using the mean value theorem on the interval $[0,x]$.

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  • $\begingroup$ I did. So $f(t) = e^{-t}, f'(t) = -e^{-t}$. So the MVT states: $\frac{f(t)-f(0)}{t-0}=f'(\xi)$ thus: $\frac{e^{-t}-1}{t-0}=f'(\xi)=-e^{-\xi}$ But what does that show me? $\endgroup$ – de_dust Jan 12 '17 at 21:14
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    $\begingroup$ Well, $-e^{-\xi}\geq -1$ since $\xi\geq 0$, so you get $\frac{e^{-x}-1}{x}\geq -1$, or $e^{-x}\geq 1-x$. $\endgroup$ – carmichael561 Jan 12 '17 at 21:22
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Let $f(x) = e^{-x} - (1 - x)$. So, $f '(x) = -e^{-x} + 1$.

Suppose $x \ge 0$. Observe that $f '(x) \ge 0$ (since $-1 \le -e^{-x} < 0$ for $x \ge 0$). Since $f '(x) \ge 0$ on the interval $[0, x]$ for $x \ge 0$, $f$ is increasing on this interval. Thus, $f(x) \ge f(0) = 0$ for $x \ge 0$. Therefore, $e^{-x} - (1 - x) \ge 0 \implies e^{-x} > 1 - x$ for $x > 0$ (i.e: on the interval $(0, \infty)$).

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Suppose $x>0$. Then the mean value theorem applied to the interval $[0,x]$ says that $$ \frac{e^{-x}-e^{-0}}{x-0}=-e^{-c} $$ for some $c\in(0,x)$. Since $c>0$, we have $-c<0$, so $e^{-c}<1$ and so $-e^{-c}>-1$.

Therefore $$ e^{-x}-1 > -x $$

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Let $f(x)=e^{-x}-1+x$. Fix a value $b \in [0,\infty)$ and consider the interval $[0,b]$. Note that $f(0)=0$.

Now according to the Mean Value Theorem, there exist a $c\in (0,b)$ such that $$f'(c)=\frac{f(b)-f(0)}{b-0}=\frac{e^{-b}-1+b}{b}>0.$$

This shows us that $f$ is increasing on any interval $(0,b)$ where $b$ is some fixed value in $[0,\infty)$. So $f(x)\geq f(0)$ for all $x\geq 0$. Thus $e^{-x}\geq 1-x$ for all $x\geq 0$.

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