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: $\cosh^2x - \sinh^2x=1$

Note $$\cosh^2 x =1+\sinh^2 x \ge 1$$
So $\cosh^2 x \ge 1$. Since $\cosh x$ is positive, we have $$\cosh x \ge 1$$
Done! You could also say $$\cosh x =\frac{e^x+e^{-x}}{2} \ge \sqrt{e^{x} \times e^{-x}}=1$$
From $\text{AM-GM}$. 
