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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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    updated the question with the defenition.2012-04-28

3 Answers 3

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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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    @RB14: Yes, that was a slip. Thanks. It’s fixed now.2012-04-29
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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 $\exp\,x \geq 1+x$.2012-04-29
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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 $p$refer not to use derivatives, but that's a nice answer!2012-04-28