(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."

$$h(x) = x \log_2\frac{1}{x}+(1-x) \log_2\frac{1}{1-x}\ge1- \left(1-\frac{x}{1-x}\right)^2$$
If we let $x=\frac{1+y}{2}$ and push through the algebra, the claim is equivalent to:
$$1-\frac{1}{2}\log_2(1-y^2)-\frac{y}{2}\log_2\left(\frac{1+y}{1-y}\right)\ge1-\frac{4y^2}{(1-y)^2}$$
which can be rearranged to:
$$\frac{8y^2}{(1-y)^2}\ge \log_2(1-y^2) + y\log_2\left(\frac{1+y}{1-y}\right)$$
Now, using the well-known $\ln u\le u-1$:
$$\ln v^2\le v^2-1\implies \ln v \le \frac{v^2-1}{2}$$
and taking $v=\frac{1+y}{1-y}$, this becomes:
$$\ln\left(\frac{1+y}{1-y}\right)\le \frac{2y}{(1-y)^2}$$
Noting that $\log_2(1-y^2)\le0$, we can see
$$\log_2(1-y^2) + y\log_2\left(\frac{1+y}{1-y}\right)\le y\frac{1}{\ln2}\frac{2y}{(1-y)^2}=\frac{2}{\ln2}\frac{y^2}{(1-y)^2}\le \frac{8y^2}{(1-y)^2}$$
and we are done for $y>0$. 
For $y<0$ it's a little trickier, because the other $\log$ term does come into play - still working out a nice argument for how to bound things appropriately.
