how to prove $x^2\ln{x}+x+e^x-3x^2>0$

Question

Let $x>0$; show that: $$f(x)=x^2\ln{x}+x+e^x-3x^2>0$$

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

Answer 1

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 $$e^x-\frac{x^2-2x}{x^2-4x+6}>1+x-\frac{x^2-2x}{x^2-4x+6}=\frac{x^3-4x^2+4x+6}{x^2-4x+6}>0.$$

Answer 2

$$x^2-x+e^x(x-2)=0 \implies e^{-x}=\frac{x-2}{x(1-x)}$$ 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 $1.768$.

Answer 3

It suffices to prove that for $x > 0$, $$x\ln x + 1 + \tfrac{1}{x}\mathrm{e}^x - 3x > 0.$$ Denote LHS by $f(x)$. We have, for $x > 0$, $$f'(x) = \ln x + \tfrac{1}{x}\mathrm{e}^x - \tfrac{1}{x^2}\mathrm{e}^x - 2$$ and $$f''(x) = \frac{1}{x} + \frac{1}{x^3}\mathrm{e}^x(x^2 - 2x + 2) > 0.$$ Thus, $f'(x)$ is strictly increasing on $(0, \infty)$. Since $f'(1.763) < 0$ and $f'(1.764) > 0$, the equation $f'(x) = 0$ has a unique positive real root $x_0\in (1.763, 1.764)$. Thus, $f(x)$ is strictly decreasing on $(0, x_0)$, and strictly increasing on $(x_0, \infty)$. Thus, $f(x) \ge f(x_0)$ for $x > 0$.

Let $a = 1.763, b = 1.764$. Since $a \le x_0 \le b$, we have $f(x_0) = x_0\ln x_0 + 1 + \tfrac{1}{x_0}\mathrm{e}^{x_0} - 3x_0 \ge a\ln a + 1 + \tfrac{1}{b}\mathrm{e}^a - 3b >0 $. We are done.

Answer 4

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!