(Elegant) proof of : $x \log_2\frac{1}{x}+(1-x) \log_2\frac{1}{1-x} \geq 1- (1-\frac{x}{1-x})^2$ I am looking for the most concise and elegant proof of the following inequality:
$$
h(x) \geq 1- \left(1-\frac{x}{1-x}\right)^2, \qquad \forall x\in(0,1)
$$
where $h(x) = x \log_2\frac{1}{x}+(1-x) \log_2\frac{1}{1-x}$ is the binary entropy function. Below is a graph of the two functions.

Of course, an option would be to differentiate $1,2,\dots,k$ times, and study the function this way — it may very well work, but is not only computationally cumbersome, it also feels utterly inelegant. (For my purposes, I could go this way, but I'd rather not.)
I am looking for a clever or neat argument involving concavity, Taylor expansion around $1/2$, or anything — an approach that would qualify as "proof from the Book."

Here is a definite integral approach.
Using the known identity (for all $u > -1$)
$$\ln (1 + u) = \int_0^\infty \frac{1 - \mathrm{e}^{-us}}{s}\mathrm{e}^{-s}\,\mathrm{d} s, \tag{1}$$
we have
\begin{align*}
 &-\frac{x\ln x}{\ln 2} - \frac{(1 - x)\ln (1 - x)}{\ln 2} \\
 =\,& \int_0^\infty \left[x(\mathrm{e}^{s(1 - x)} - 1) + (1 - x)(\mathrm{e}^{sx} - 1)\right]\frac{\mathrm{e}^{-s}}{s\ln 2}\,\mathrm{d} s\\
 =\,& \int_0^\infty x(1 - x)\left(\frac{\mathrm{e}^{s(1 - x)} - 1}{1 - x} + \frac{\mathrm{e}^{sx} - 1}{x}\right)\frac{\mathrm{e}^{-s}}{s\ln 2}\,\mathrm{d} s\\
 =\,& \int_0^\infty x(1 - x)\left(\int_0^s (\mathrm{e}^{t(1-x)} + \mathrm{e}^{tx}) \,\mathrm{d} t\right)\frac{\mathrm{e}^{-s}}{s\ln 2}\,\mathrm{d} s \\
 \ge\,& \int_0^\infty x(1 - x)\left(\int_0^s 2\sqrt{\mathrm{e}^{t(1-x)} \cdot \mathrm{e}^{tx}} \,\mathrm{d} t\right)\frac{\mathrm{e}^{-s}}{s\ln 2}\,\mathrm{d} s \\
 =\,& \int_0^\infty x(1 - x)\left(\int_0^s 2\mathrm{e}^{t/2} \,\mathrm{d} t\right)\frac{\mathrm{e}^{-s}}{s\ln 2}\,\mathrm{d} s\\
 =\,& \int_0^\infty x(1 - x)(4\mathrm{e}^{s/2} - 4)\frac{\mathrm{e}^{-s}}{s\ln 2}\,\mathrm{d} s\\
 =\,& 4x(1 - x) \tag{2}
\end{align*}
where in (2) we have used $\ln (1 - 1/2) = \int_0^\infty \frac{1 - \mathrm{e}^{s/2}}{s}\mathrm{e}^{-s}\,\mathrm{d} s$ (simply letting $u = -1/2$ in (1)).
Also, we have
$$4x(1 - x) - \left[1- \left(1-\frac{x}{1-x}\right)^2\right] = \frac{x(2 - x)(1 - 2x)^2}{(1 - x)^2} \ge 0.$$
We are done.
