I need to solve the following logarithmic inequality:
Show that $\ln(x)\geq1-\frac{1}{x}$ when $x>0$
My attempt
$$\ln(x)\geq1-\frac{1}{x}$$ $$\ln(x)\geq\ln(e)-\ln(e^{\frac{1}{x}})$$ $$\ln(x)\geq\ln(\frac{e}{e^{\frac{1}{x}}})$$ $$x\geq\frac{e}{e^\frac{1}{x}}$$ $$x{e^\frac{1}{x}}\geq e$$
I think I could do the last inequality by solving for $x$ in $x{e^\frac{1}{x}}=e$ (which gives $1$ according to WolframAlpha) and then checking all three cases. The problem is that I haven't officially learned how to solve exponential equations and I was wondering if there is another way to solve this.