Jtdylktuy

I wanna show that for $a$ and $b >0$ and $<1$. Then
$$2(a-b)^2leq (a) log (a) + (1-a)log(1-a) -(a)log(b)-(1-a)log(1-b).$$
If I wanna go from left to right, then I think I gotta use Taylor's but I don't know how. Also, the right hand side I figured out is equal to
$$(a)logfrac ab+(1-a)logfrac{1-a}{1-b}.$$

As already mentioned, this is a special case of Pinsker's inequality.
The following proof for this special case is taken from Pinsker’s inequality and its applications to lower bounds:

For fixed $a in (0, 1)$ define $f:(0, 1) to Bbb R$ as
$$
f(b) = alogfrac ab+(1-a)logfrac{1-a}{1-b} - 2(a-b)^2 , .
$$
Then
$$
f'(b) = -frac ab + frac{1-a}{1-b} + 4(a-b) = (b-a) left(frac{1}{b(1-b)} - 4right) , .
$$
The second factor is $ge 0$, with equality only at a single point ($b= frac 12$), so that $f$ is strictly decreasing on $(0, a]$ and
strictly increasing on $[a, 1)$. It follows that
$$
f(b) ge f(a) = 0 , ,
$$
which is the desired inequality. Equality holds only if and only if $b=a$.

You are trying to prove Pinsker's inequality. Since both $a$ and $b$ are between 0 and 1, we can think of $(a,1-a)$ and $(b,1-b)$ as binary probability distributions. Let me call them $P$ and $Q$ respectively, so that $P(0)=a$, $P(1)=1-a$, $Q(0)=b$ and $Q(1)=1-b$.

Now it is easy to see that your left hand side is $2delta(P,Q)^2$, where $delta$ is the total variation distance between $P$ and $Q$, while the right hand side is the relative entropy, or Kullback-Leibler divergence, $D(P||Q)$.

You can find the proof of Pinsker's inequality in many places, for example here: Proof of Pinsker's inequality.