Trying to find the intervall onto which cosh(x) is strictly increasing

Question

I tried to find the interval onto which cosh(x) is strictly increasing: \begin{array}{c} cosh(x) < cosh(x+1) \\ \frac{1}{2}(e^x+\frac{1}{e^x})<\frac{1}{2}(e^{x+1}+\frac{1}{e^{x+1}})\qquad |*2\\ e^x+\frac{1}{e^x}<e^{x+1}+\frac{1}{e^{x+1}} \qquad | -e^{x+1}-\frac{1}{e^{x+1}}\\ e^x-e^{x+1}+\frac{1}{e^x}-\frac{1}{e^{x+1}}<0\\ e^x(1-e)+\frac{1}{e^x}(1-\frac{1}{e})<0 \qquad |*e^x\\ e^{2x}(1-e)+(1-\frac{1}{e})<0\\ e^{2x}(1-e)+1-\frac{1}{e}<0\qquad |-1+\frac{1}{e}\\ e^{2x}(1-e)<\frac{1}{e}-1 \qquad |:(1-e)\\ e^{2x}>(\frac{1}{e}-1)\frac{1}{1-e} = \frac{1-e}{e} \frac{1}{1-e} = \frac{1}{e}=e^{-1}\\ 2x>-1\qquad |:2\\ x>-\frac{1}{2} \end{array} But this is obviously wrong, it should be: \begin{array}{c} x>0\\ \end{array} Where is my mistake?