Question

Prove that cosh x >= 1 for all x

Answer 1

if it helps:
\[\cosh(x) ={1/2}*(e^x + e ^{-x})\]

Answer 2

We can apply AM-GM:\[ (e^x/2+e^{-x}/2)/2\ge \sqrt(e^x*e^{-x}/4)=1/2 \rightarrow (e^x/2+e^{-x}/2)\ge 1\] Then, we note that if x is negative, e^x becomes e^-x and e^-x becomes e^x, so we get the same expression.