(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."

Hopefully, this is right! 
Note that from the weighted AM-GM inequality, We have that  $$h(x)=\log_2{\frac{1}{x^x(1-x)^{1-x}}} \ge \log_2\frac{1}{x^2+(1-x)^2}$$
Thus we have to show $$\left(1-\frac{x}{1-x}\right)^2 \ge 1-\log_2\frac{1}{2x^2-2x+1}=\log_2{(4x^2-4x+2)}$$ Substitute $x=\frac{a+1}{a+2}$, and we have $$f(a)=a^2 -\log_2\left(\frac{a^2}{(a+2)^2}+1 \right) \ge 0$$
For $a \ge -1$. Differentiating gives $$f'(a)=2a\left(1-\frac{1}{(a+2) (a^2+2 a+2) \log(2)}\right)$$ and this $f'(a)>0$ for $a>0$ alternatively $f(a) \ge f(0)=0$ for $a \ge 0$. 
Also,  notice the local maxima lies between $-1$ and $0$.  But since $f(-1)=f(0)=0$, we have that $f(a) \ge 0$ for $-1 \le a \le 0$.  
Our proof is done. 
