I think it's pretty easy question, maybe even dumb one, but still I can't find a nice way to solve it.

How do you prove that $\forall x . \cosh(x) \ge 1$ , without using the identity: $\cosh^2x-\sinh^2x=1$, and not without using derivatives.

the defenition of $\cosh(x)$ is: $\frac{e^x+e^{-x} }{2}$

I already proved that using the identity, but I'm just wondering if there is another way. sadly, I couldn't find a proof using the search engine, even though that is a very basic question.

would appreciate your help!

  • 4
    $\begingroup$ What is the smallest x + 1/x can get? Definition of Cosh? $\endgroup$ – Chris K. Caldwell Apr 28 '12 at 22:42
  • $\begingroup$ How you prove this (and, really, whether you have proved this) depends sensitively on how you define $\cosh$ and what you assume known about the things you have used to define $\cosh$. What is your definition of $\cosh$, and what do you assume known? $\endgroup$ – leslie townes Apr 28 '12 at 22:43
  • $\begingroup$ updated the question with the defenition. $\endgroup$ – rboy Apr 28 '12 at 22:50

You have by definition that $$\cosh x=\frac12(e^x+e^{-x})\;,$$ so it suffices to show that $e^x+e^{-x}\ge 2$. But this is an easy consequence of the fact that for any positive real number $u$, $u+u^{-1}\ge 2$. To see this, note that the desired inequality is equivalent to $$\frac{u^2+1}u\ge 2$$ and hence to $u^2+1\ge 2u$, or $u^2-2u+1\ge 0$. But this is clearly true, since $$u^2-2u+1=(u-1)^2\;.$$ Reorganizing in logical order: for $u\ne 0$,

$$\begin{align*} u^2-2u+1=(u-1)^2\ge 0&\implies u^2+1\ge 2u\\ &\implies\frac{u^2+1}u\ge 2\\ &\implies u+\frac1u\ge 2\;, \end{align*}$$

and in particular this holds when $u=\cosh x$ for any real $x$, since it’s clear from the definition that $\cosh x>0$ for all $x$.

  • 1
    $\begingroup$ lovely, I knew it must be an easy way! thanks a lot! $\endgroup$ – rboy Apr 28 '12 at 22:54
  • 1
    $\begingroup$ @RB14 If you know the AM-GM inequality, your inequality $u + u^{-1} \geq 2$ comes almost immediately from this. $\endgroup$ – user38268 Apr 29 '12 at 0:04
  • $\begingroup$ actually, in second reading, you did make some mistake here. you said that the inequality $u+u^{-1}\ge 2$ is true for any non-zero real number $u$, but actually it's not true for negative numbers, so it's true only for any real number greater than 0. but it's still good for proving the expression since $e^x > 0$. $\endgroup$ – rboy Apr 29 '12 at 14:08
  • 1
    $\begingroup$ @RB14: Yes, that was a slip. Thanks. It’s fixed now. $\endgroup$ – Brian M. Scott Apr 29 '12 at 14:19

Having $\small e^x=1+x+x^2/2!+x^3/3! +x^4/4! +... $ we have also $$\small \cosh(x)={(e^x+e^{-x})\over 2}=1 + x^2/2! + x^4/4! + \ldots \ge 1 $$ for all real x just by adding the formal representation of the series for x and for -x termwise, where the terms at odd exponents of x vanish because of the alternating sign.

  • 3
    $\begingroup$ Even easier: one can start with the basic inequality $\exp\,x \geq 1+x$. $\endgroup$ – J. M. is a poor mathematician Apr 29 '12 at 0:57

By definition, $\cosh x=\frac{e^x+e^{-x}}{2}$. Taking the derivative, we get $\frac{e^x-e^{-x}}{2}$, which is $0$ when $x=0$, positive when $x>0$ and negative when $x<0$. So $x=0$ is a the function's minimum, and $\cosh 0=1$.

  • 1
    $\begingroup$ Thanks for your answer, but I forgot to mention that I prefer not to use derivatives, but that's a nice answer! $\endgroup$ – rboy Apr 28 '12 at 22:44

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.