${{x}^{\ln x}}+{{e}^{{{\ln }^{2}}x}}\le 2{{e}^{4}}\left( * \right)$

Ta có: ${{e}^{{{\ln }^{2}}x}}={{\left( {{e}^{\ln x}} \right)}^{\ln x}}={{x}^{\ln x}}$

Vậy $\left( * \right)$

$\begin{align} & \Leftrightarrow 2{{e}^{{{\ln }^{2}}x}}\le 2{{e}^{4}}\Leftrightarrow {{\ln }^{2}}x\le 4\Leftrightarrow \left| \ln x \right|\le 2 \\ & \Leftrightarrow -2\le \ln x\le 2\Leftrightarrow {{e}^{-2}}\le x\le {{e}^{2}}\Leftrightarrow \frac{1}{{{e}^{2}}}\le x\le {{e}^{2}} \\ \end{align}$