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I have a function in the shape $1+x$ (for some value of $x$). I have used the standard inequality to get an upper bound of $e^x$. Now, I need to get a lower bound. I have found an inequality $\left(1+x\right)^{n} \geq e^{x}\left(1-x^{2}\right)$. However, I then exponentiate the bonds, which in the case of the upper bound gives me $e^{nx}$ but the lower bound turns out to be quite ugly. Perhaps a bound of the shape $1+x \geq e^{f(x)}$ for some function $f$ could hold?

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    $\begingroup$ maybe $f(x) = ln(x)$ ? $\endgroup$
    – infinity
    Feb 23 '20 at 10:00
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    $\begingroup$ @user2316602 It's impossible. Try $x=-1$. For $x>-1$ there is $f(x)=\ln(1+x).$ $\endgroup$ Feb 23 '20 at 10:12
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As correctly pointed out in the comments, I can use $\ln(1+x)$ and clearly, nothing "better" can exist.

However, I have found the inequality $1-x \geq e^{-\frac{x}{1-x}}$ which holds for $x \leq 1$. This is very useful and I have used it several times. Often, one has something like $1-1/n$, which is very close to $e^{-1/n}$. However, one often needs and actual bound, not that it is asymptotically equivalent. In that case, the mentioned bound is very useful.

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