How do I continue to prove this?
Show that $$ x \ln(ex) - \sqrt{x}\geq 0 $$ for all $$ x \geq1 $$
My try:
$$\begin{eqnarray*} \\ \ln(e^x) + \ln(x^x) &\geq& \sqrt{x} \\ \\ \ln(e^x) &\geq& \sqrt{x} - \ln(x^x) \\ \\ e^x &\geq & e^\sqrt{x} e^{-\ln(x^x)} \\ \\ e^x&\geq& e^\sqrt{x} (1/x^x) \\ \\ e^x - \frac{e^\sqrt{x}}{x^x}&\geq& 0 \end{eqnarray*}$$
I "see" that this is bigger than 0, but I think that there are more calculations to do.