# Inequality $1+x \geq e^{f(x)}$?

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?

• maybe $f(x) = ln(x)$ ? Feb 23 '20 at 10:00
• @user2316602 It's impossible. Try $x=-1$. For $x>-1$ there is $f(x)=\ln(1+x).$ Feb 23 '20 at 10:12

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.