A tough inequality After fiddling with things, I came to wonder whether the following holds for $0<x<1:$
$$\left(\frac{1}{x}-1\right)^x \geq (x-1)(4x^2-4x-1)$$
To get an idea of what I was dealing with I plotted the curves, and it turns out these are very close, especially on $(\frac{1}{2},1)$, but it appears to be true on the entire interval. Any ideas how one could prove this?

The Taylor series for $(\frac 1x-1)^x$ at center $1/2$, is $$1-2(x-1/2)-2(x-1/2)^2+4(x-
1/2)^3+2/3(x-1/2)^4+\cdots$$
The RHS, namely $(x-1)(4x^2-4x-1)$, is the third-degree Taylor approximation to the LHS at 0, i.e. the Taylor series cut off after the cubic term.  Unfortunately, the series is not alternating, so I don't have an immediate proof that rounding to the cubic approximation will always decrease the value, so this doesn't directly answer the inequality question posed (although it does explain why the two are very close).
