show that $\cosh x \geq1$ Can someone help me show that $\cosh \geq1$. 
I know that $\cosh x = \frac{1}{2}(e^x+s^{-x})$
I think I'm supposed to use the identity: $\cosh^2x - \sinh^2x=1$

by AM-GM we get : $$\frac{1}{2}(e^x+e^{-x})\geq \sqrt{e^x\cdot e^{-x}}=1$$
