A function inequality about $e^x$ and $\ln x$

Question

If for any $x \in (1,+\infty)$，there is the inequality: $$x^{-3} e^{x}-a \ln x \geq x+1$$ Find the value range of $a$ .

And I tried constructing the function $f(x)=x^{-3} e^{x}-a \ln x - x-1$ and deriving it, but to no avail.

Answer 1

$$\max{a}=\min_{x>1}\frac{\frac{e^x}{x^3}-x-1}{\ln{x}}.$$ Let $g(x)=\frac{\frac{e^x}{x^3}-x-1}{\ln{x}}.$

Thus, for $e^x=x^3$ or $x=3\ln{x}$ we obtain: $$g'(x)=\frac{x^4+x^3+e^x(x\ln{x}-3\ln{x}-1)-x^4\ln{x}}{x^4\ln^2x}=$$ $$=\frac{x^4+x^3+x^3(x\ln{x}-3\ln{x}-1)-x^4\ln{x}}{x^4\ln^2x}=\frac{x^4-3x^3\ln{x}}{x^4\ln^2x}=0.$$ Also, for $e^x=x^3$ we obtain: $$\frac{\frac{e^x}{x^3}-x-1}{\ln{x}}=-\frac{x }{\ln{x}}=-3.$$ Now, we'll prove that $$\frac{\frac{e^x}{x^3}-x-1}{\ln{x}}\geq-3$$ or $$\frac{e^x}{x^3}-x-1+3\ln{x}\geq0,$$ where the minimum occurs for $x=x_0$, where $x_0>1$ is a root of the equation: $e^x=x^3$.

Indeed, let $h(x)=\frac{e^x}{x^3}-x-1+3\ln{x}.$

Thus, $$h'(x)=\frac{e^x}{x^3}-\frac{3e^x}{x^4}-1+\frac{3}{x}=\frac{(x-3)(e^x-x^3)}{x^4}.$$ Now, the equation $e^x=x^3$ or $x=3\ln{x}$ has two roots maximum because $\ln$ is a concave function.

But $3\ln3-3>0$, $3\ln1-1<$ and $3\ln5-5<0$ which says that the equation $e^x=x^3$ has two roots $1<x_0<3$ and $x_1>3$ exactly and easy to see that $x_{max}=3$ and $x_{min_1}=x_0$, $x_{min_2}=x_1.$

Thus, $$h(x)\geq h(x_0)=h(x_1)=0$$ and we got the answer: $$(-\infty,-3]$$

Answer 2

Since $x = \ln(x^{-3}\mathrm{e}^x) + 3\ln x$ for all $x > 1$, we have \begin{align} \frac{x^{-3}\mathrm{e}^x - x - 1}{\ln x} &= \frac{x^{-3}\mathrm{e}^x - (\ln(x^{-3}\mathrm{e}^x) + 3\ln x) - 1}{\ln x}\\ &= \frac{x^{-3}\mathrm{e}^x - 1 - \ln(x^{-3}\mathrm{e}^x)}{\ln x} - 3. \end{align} Note that $u - 1 - \ln u \ge 0$ for all $u > 0$, with equality only at $u=1$. Thus, the minimum of $\frac{x^{-3}\mathrm{e}^x - x - 1}{\ln x}$ on $(1, \infty)$ is $-3$ achieved at $x^{-3}\mathrm{e}^x = 1$ (it is easy to prove that it has two real solutions on $(1, \infty)$).