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?
 A: As suggested by Karl Kronenfield, set $f(x) = x^x (1-x)^{1-x} (1+4x-4x^2)$ and set $g(x) = \log f(x)$. The goal is to prove $f(x)  \leq 1$ or, equivalently, $g(x) \leq 0$. As pointed out by Karl, $f(x) = f(1-x)$, so it suffices to prove the inequality on $[0,1/2]$.
Below, I plot $g$ (red), $g'$ (blue) and $g''$ (green) on $(0,1/2)$ for illustrative purposes; this proof does not rely on the picture. (The vertical scales for the three graphs are different.)

Note that
$$\frac{d^2}{(dx)^2} g(x) = - \frac{(1 - 2 x)^2 (1 - 12 x + 12 x^2)}{x (1 - x)  (1 + 4 x - 4 x^2)^2}$$
From this we can compute that $g''(x)$ is well defined on $(0,1/2)$, vanishing at $1/2$ and at $\alpha := (3-\sqrt{6})/6 \approx 0.092$, and vanishing nowehere else on $(0,1/2)$.  We see that $g''(x)$ is positive on $(0, \alpha)$ and negative on $(\alpha, 1/2)$. 
So $g'$ is increasing on $(0, \alpha)$ and decreasing on $(\alpha, 1/2)$. Combined with the facts that $g'(0) = - \infty$ and $g'(1/2)=0$, we see that $g'$ must be positive on $(\alpha, 1/2)$ and have a single zero in $(0, \alpha)$. Denote that single zero by $\beta$. Numerically, $\beta \approx 0.039$. 
So $g$ is decreasing on $(0, \beta)$ and increasing on $(\beta, 1/2)$. In particular, $g$ assumes its largest values at the endpoints of $(0,1/2)$. By direct computation, $g(0) = g(1/2) = 0$ as desired. 

Methodology: Plot $f$ and notice that it is concave down on a large neighborhood of $1/2$, which immediately proves the inequality on that neighborhood. Decide to compute the exact size of that neighborhood. Realize that $\frac{d^2}{(dx)^2} f(x)$ is a giant mess, but $\frac{d^2}{(dx)^2} \log f(x)$ is pretty simple. Draw a sketch of $g$ with a hypothetical "spike towards positivity" in $[0,1/2]$. Think about how the concavity of that hypothetical $g$ differs from the concavity of the true $g$. Write the answer. Then rewrite the answer with added graphics and detail to sound prettier.
A: 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).
