# How to prove $\cosh(x) \ge 1$ without the $\cosh^2x-\sinh^2x=1$ identity

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.

• What is the smallest x + 1/x can get? Definition of Cosh? Apr 28, 2012 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? Apr 28, 2012 at 22:43
• updated the question with the defenition.
– rboy
Apr 28, 2012 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 $x$.

• lovely, I knew it must be an easy way! thanks a lot!
– rboy
Apr 28, 2012 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, 2012 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, 2012 at 14:08
• @RB14: Yes, that was a slip. Thanks. It’s fixed now. Apr 29, 2012 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.

• Even easier: one can start with the basic inequality $\exp\,x \geq 1+x$. Apr 29, 2012 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 $\cosh 0=1$.

• 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, 2012 at 22:44

Depending on your definition of $$\cosh x$$, we could also define $$\cosh x$$ as $$\cos(ix)$$, and now we can use Taylor series (like in Gottfried Helms's answer):

$$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{4} - \cdots$$ $$\cosh x = 1 - \frac{(ix)^2}{2} + \frac{(ix)^4}{4} - \cdots$$

When the exponent is in the form $$4n$$, we have $$i^{4n} = (i^4)^n = 1$$, and when it is in the form $$4n+2$$, we have $$i^{4n+2} = i^{4n} \cdot i^2 = -1$$, so now all the terms become positive:

$$\cosh x = 1 + \frac{x^2}{2} + \frac{x^4}{4} + \cdots ≥ 1$$

We could also use the fact that $$e^x ≥ 1 + x$$ for all real $$x$$ as mentioned in the comment by J.M. below:

$$\cosh x = \frac{e^x + e^{-x}}{2} ≥ \frac{1+x+1-x}{2} ≥1$$