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:

Prove inequality using Mean Value Theorem 2

Prove inequality using Mean Value Theorem Mean Value theorem problem?(inequality)

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


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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.