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.
would appreciate your help!
 A: 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.
A: 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 $\cosh 0=1$.
A: 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 $x$.
A: 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:
$$\cosh x = \frac{e^x + e^{-x}}{2} ≥ \frac{1+x+1-x}{2} ≥1$$
