Claim :
$$\left(1+e^{-1}\right)^{3}\cdot\left(1-e^{-1}\right)-\frac{\left(\sqrt{5}+1\right)}{2}<0$$
I found it accidentaly using Desmos .It's quite good as approximation of the golden ratio.
Some thought :
I recall that :
$$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$$
And we can also use the Binet's formula to get a good approximation of $\phi$ using the $n$ th root .
Edit :
A related result :
Let :
$$f\left(x\right)=\left(x+1\right)^{3}\left(1-x\right)-x-\frac{5}{4}$$
Then :
$$f\left(e^{-1}\right)<0\quad (1)$$
And :
$$f\left(\frac{\left(\sqrt{3}-1\right)}{2}\right)=0\quad(2)$$
Edit after the comment :
We have from $(1)$ and $(2)$ (the extrema is also a root) :
$$\left(1+e^{-1}\right)^{3}\cdot\left(1-e^{-1}\right)<e^{-1}+1.25$$
Now using :
Let :
$$I_n(x)=e^x-\sum_{i=0}^{n}\frac{x^i}{i!}$$
Then we have as $|x|\leq 1$ :
$$|I_n(x)|\leq \frac{1}{(n+1)!}\left(1+\frac{2}{n+1}\right)$$
And using again the Binet's formula the result follows .
Last edit : An inequality for the fun !
$$\left(1-\left(\left(-\frac{1}{2e-2}\right)\cdot e^{-1}+\left(\frac{1}{2e-2}+1\right)\cdot\left(\frac{\left(\sqrt{3}-1\right)}{2}\right)\right)\right)\left(1+\frac{\left(\sqrt{3}-1\right)}{2}\right)^{e}\left(1+e^{-1}\right)^{\left(3-e\right)}-\frac{\left(\sqrt{5}+1\right)}{2}<0$$
Question :
How to show it by hand ?
Thanks in advance .