how to prove $x^2\ln{x}+x+e^x-3x^2>0$ Let $x>0$; show that:
$$f(x)=x^2\ln{x}+x+e^x-3x^2>0$$
It seem this inequality 
$$f'(x)=2x\ln{x}+1+e^x-5x$$
$$f''(x)=2\ln{x}+e^x-3$$
$$f'''(x)=\dfrac{2}{x}+e^x>0$$

$$x^2-x+e^x(x-2)=0 \implies e^{-x}=\frac{x-2}{x(1-x)}$$ and the solution cen write using the generalized Lambert function (have a look at equation $(4)$).
By inspection or graphing, the positive root is close to $x=2$. To have a better value, expand the lhs as a Taylor series
$$y=x^2-x+e^x(x-2)=2+\left(3+e^2\right) (x-2)+\left(1+e^2\right) (x-2)^2+\frac{1}{2} e^2
   (x-2)^3+O\left((x-2)^4\right)$$
Use series reversion to get
$$x=2+\frac{y-2}{3+e^2}-\frac{\left(1+e^2\right)
   (y-2)^2}{\left(3+e^2\right)^3}+\frac{\left(4+5 e^2+3 e^4\right) (y-2)^3}{2
   \left(3+e^2\right)^5}+O\left((y-2)^4\right)$$ Make $y=0$ to get
$$x=\frac{2 \left(136+257 e^2+196 e^4+76 e^6+14
   e^8+e^{10}\right)}{\left(3+e^2\right)^5}\approx 1.771$$ while the exact solution is $1.768$.
