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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!

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What is the smallest x + 1/x can get? Definition of Cosh? – Chris K. Caldwell Apr 28 '12 at 22:42
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? – leslie townes Apr 28 '12 at 22:43
updated the question with the defenition. – rboy Apr 28 '12 at 22:50
up vote 12 down vote accepted

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 $x$.

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lovely, I knew it must be an easy way! thanks a lot! – rboy Apr 28 '12 at 22:54
@RB14 If you know the AM-GM inequality, your inequality $u + u^{-1} \geq 2$ comes almost immediately from this. – user38268 Apr 29 '12 at 0:04
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 $e^x > 0$. – rboy Apr 29 '12 at 14:08
@RB14: Yes, that was a slip. Thanks. It’s fixed now. – 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.

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Even easier: one can start with the basic inequality $\exp\,x \geq 1+x$. – J. M. 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 $\cosh 0=1$.

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

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