I have trouble with proving the following inequality
$$\frac{x}{e^x}<\frac{1.5}{x^2}$$ for $x>0$.
My attempt: $$1.5e^x-x^3>0$$ Let $f(x)=1.5e^x-x^3$ $$f'(x)=1.5e^x-3x^2 $$ $$f'(x)=0$$ $$e^x=2x^2$$ Using the Lambert $W$-function I get three solutions: $$x=-0.539835, x=1.48796,x=2.61787$$ Drawing the graph of $f'(x)$ I can see that the derivative is not always greater than $0$ for $x>0$, so I cannot say the function monotonically increases.
Is there a simpler way to prove this? Maybe using the Lambert $W$-function is not necessary at all?
Because $x^2$ is positive, proving this inequality is equivalent to proving that
$$\frac{x^3}{e^x} < 1.5$$
Let's call $f(x) = x^3/e^x$. Then
$$f'(x) = \frac{3x^2}{e^x} - \frac{x^3}{e^x} = \frac{(3-x)x^2}{e^x}$$
Note that for $x>0$, $f'(x)$ has a unique root, $x=3$. For $0<x<3$, $f'(x)>0$, and for $x>3$, $f'(x)<0$. Thus, $f(x)$ has a unique global maximum at $x=3$. For all $x>0$, $f(x) \le f(3)$.
$$f(3) = \frac{27}{e^3} \approx 1.34425 < 1.5$$
Thus, the inequality is true.