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 
$$f'(x)=2x\ln{x}+1+e^x-5x$$
$$f''(x)=2\ln{x}+e^x-3$$
$$f'''(x)=\dfrac{2}{x}+e^x>0$$

We need to prove that $$\ln{x}+\frac{1}{x}+\frac{e^x}{x^2}-3>0$$ for $x>0$.
But $$\left(\ln{x}+\frac{1}{x}+\frac{e^x}{x^2}-3\right)'=\frac{x^2-x+e^x(x-2)}{x^3}$$ increases for $x>0$ because easy to show that 
$$\left(\frac{x^2-x+e^x(x-2)}{x^3}\right)'=\frac{(x^2-4x+6)e^x-(x^2-2x)}{x^4}>0,$$
which says that the minimum occurs for the unique positive root of the equation:
$$x^2-x+e^x(x-2)=0.$$
The inequality $(x^2-4x+6)e^x-(x^2-2x)>0$ we can prove by the following way:
We need to prove that:
$$e^x>\frac{x^2-2x}{x^2-4x+6},$$ which is true because
$$e^x-\frac{x^2-2x}{x^2-4x+6}>1+x-\frac{x^2-2x}{x^2-4x+6}=\frac{x^3-4x^2+4x+6}{x^2-4x+6}>0.$$
