<s>[INST] 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 
$$f'(x)=2x\ln{x}+1+e^x-5x$$
$$f''(x)=2\ln{x}+e^x-3$$
$$f'''(x)=\dfrac{2}{x}+e^x>0$$
 [/INST] By my previous post it's enough to prove that:
$$x^2\ln{x}+x+e^x-3x^2>0,$$ where $x$ is a root of the equation $$x^2-x+e^x(x-2)=0.$$
The last equality gives $$e^x=\frac{x^2-x}{2-x}$$ or
$g(x)=0,$ where $$g(x)=x-\ln\frac{x^2-x}{2-x}$$  and since $$\frac{x^2-x}{2-x}>0,$$ we obtain $$1<x<2.$$
Now, $$g'(x)=\left(x-\ln\frac{x^2-x}{2-x}\right)'=-\frac{x^3-4x^2+6x-2}{x(2-x)(x-1)}=-\frac{x(x-2)^2+2(x-1)}{x(2-x)(x-1)}<0,$$
which says that $g$ decreases.
Also, $$g(1.7672)=0.0052...>0,$$ which says that the root of the equation $x^2-x+e^x(x-2)=0$ is greater than $1.7672$.
Id est, it's enough to prove that $$x^2\ln{x}+x+e^x-3x^2>0,$$ where $e^x=\frac{x^2-x}{2-x}$ and $1.7672<x<2$ for which it's enough to prove that:
$$x^2\ln{x}+x+\frac{x^2-x}{2-x}-3x^2>0$$ or $f(x)>0$, where 
$$f(x)=\ln{x}-\frac{6x-3x^2-1}{x(2-x)}.$$
But $$f'(x)=\frac{x^3-4x^2+6x-2}{(x-2)^2x^2}=\frac{x(2-x)^2+2(x-1)}{x^2(2-x)^2}>0,$$
which says that $f$ increases.
Thus, $$f(x)>f(1.7672)=0.000096...>0$$ and we are done!
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