\[\text{please prove}\quad\forall x>0,\quad xe^x-\ln{x}>\frac{3}{2}\tag{@} \]

\[(@)\Leftrightarrow f(x)=\frac{e^x}{x^a}>\frac{\ln{x}+\frac{3}{2}}{x^{a+1}}=g(x)\\ f'(x)=\frac{e^x(x-a)}{x^{a+1}}\\ g'(x)=\frac{1-(a+1)(\ln{x}+\frac{3}{2})}{x^{a+2}}\\ \because\text{we want}\quad f(x)_{min}>g(x)_{max}\therefore a>0\\ \therefore f(x)_{min}=f(a)=\frac{e^a}{a^a}\\ \therefore g(x)_{max}=g(e^{\frac{1}{a+1}-\frac{3}{2}})=\frac{e^{\frac{3}{2}a+\frac{1}{2}}}{a+1}\\ \text{then we want}\quad \frac{e^a}{a^a}>\frac{e^{\frac{3}{2}a+\frac{1}{2}}}{a+1}\\ \Leftrightarrow \ln{(a+1)}-a\ln{a}-\frac{a+1}{2}=h(a)>0\\ h'(a)=\frac{1}{a+1}-\ln{a}-\frac{3}{2}\quad\big\downarrow\\ h(0)->-\frac{1}{2},h'(0)->+\infty\\ \text{try}\quad a=1,\quad h(1)=\ln{2}-1<0,h'(1)=-1\\ \text{try}\quad a=\frac{1}{2},\quad h(\frac{1}{2})=\ln{3}-\frac{\ln{2}}{2}-\frac{3}{4}=\frac{1}{4}\ln{(\frac{81}{4e^3})}>0\\ \therefore a=\frac{1}{2}\quad \text{is good} \]

\[(@)\Leftrightarrow\frac{e^x}{\sqrt{x}}=f(x)>g(x)=\frac{\ln{x}+\frac{3}{2}}{x\sqrt{x}}\\ f'(x)=\frac{e^x(x-\frac{1}{2})}{x\sqrt{x}}\\ g'(x)=\frac{-\frac{5}{4}-\frac{3}{2}\ln{x}}{x^2\sqrt{x}}\\ \therefore f(x)_{min}=f(\frac{1}{2})=\sqrt{2e},\quad g(x)_{max}=g(e^{-\frac{5}{6}})=\frac{2}{3}e^{\frac{5}{4}}\\ \therefore f(x)\ge f(x)_{min}>g(x)_{max}\ge g(x) \]