# Prove the following inequality using Mean Value Theorem

First off, I've seen a couple questions similar to this one (different inequalities, same principle) but didn't really understand the answers. Here are a couple of those questions:

$$1 + 2x < e^{2x} < (1-2x)^{-1}, \ \forall \ \textrm{x} \in \ \ ] 0,1/2 [$$

When using the Mean Value Theorem to prove inequalities, remember the conclusion of the MVT:
$$\frac{f(b)-f(a)}{b-a} = f'(t)$$ for some $t$ between $a$ and $b$. Replacing $b$ by a variable $x$, and applying some algebra, we get $$f(x) = f(a) + f'(t)(x-a)$$ The case $a=0$ is particularly useful; it says: $$f(x) = f(0) + f'(t)x$$ for some $t$ with $0 < t < x$. If you can give upper and/or lower bounds for $f'(t)$, then you have an equality for $f(x)$ in terms of $x$.

Your example suggests $f(x) = e^{2x}$. Since $f'(t) = 2e^{2t}$, and $e^{2t} \geq 1$ for all $t\geq 0$, we know $f'(t) \geq 2$. So $e^{2x} > 1 + 2x$.

What about the other part of the inequality? $e^{2x} < (1-2x)^{-1}$ doesn't look like it fits the pattern above. But again with some algebra, $$e^{2x} < \frac{1}{1-2x} \implies e^{-2x} > 1 - 2x$$ and now you might see how to adapt the previous case.

• Can you explain the trick to switch the e^-2x with (1 / 1 - 2x) – Edgar Aroutiounian Jun 28 '18 at 16:14
• I got it now, and was able to do the second one. Thanks! – chilliefiber Jun 28 '18 at 16:53