Let $0\leq p \leq 1$ and $\phi(t)=t\log \frac{t}{p} + (1-t)\log \frac{1-t}{1-p}$. Prove $\phi(t)\geq 2(t-p)^2$ for $t\in[0,1]$.
Here's how I started. $$\phi'(t) = -\log \frac{1-t}{1-p}+\log\frac{t}{p}$$ $$\phi''(t)=\frac{1}{t(1-t)}$$ $$\phi'''(t)=\frac{2t-1}{t^2(1-t)^2}$$
Now, by Taylor series about $p$, $$\phi(t) = \frac{1}{p(1-p)}\frac{(t-p)^2}{2}+\frac{2c-1}{c^2(1-c)^2}\frac{(t-p)^3}{6}\geq (t-p)^2+\frac{2c-1}{c^2(1-c)^2}\frac{(t-p)^3}{6}$$ for some $c$ between $p$ and $t$ (since $\frac{1}{p(1-p)}\geq 2$).
But I'm stuck there. How do I get a coefficient 2 in front of $(t-p)^2$ and get rid of the remainder term?
I verified that the inequality is true by doing a plot. Here's the matlab code:
a = linspace(0.01,0.99,100);
[t,p] = meshgrid(a,a);
s = t.*log(t./p)+(1-t).*log((1-t)./(1-p))-2*(t-p).^2;
Use a second-order Taylor approximation with the Lagrangian remainder. Since $\phi(p) = \phi'(p) = 0$, we have
$$\phi(t) = \phi(p) + \phi'(p)(t-p) + \frac{1}{2} \phi''(c)(t-p)^2 = \frac{1}{2}\phi''(c)(t-p)^2 = \frac{(t-p)^2}{2c(1-c)}$$
for some $c$ between $p$ and $t$. Now we have $c \in (0,1)$, since $t\in [0,1]$, and the function $x \mapsto x(1-x)$ attains its maximum $\frac{1}{4}$ on the interval $[0,1]$ at the point $x = \frac{1}{2}$. Hence $0 < c(1-c) \leqslant \frac{1}{4}$, and
$$\frac{1}{2c(1-c)} \geqslant 2,$$
which yields the desired
$$\phi(t) = \frac{1}{2c(1-c)}(t-p)^2 \geqslant 2(t-p)^2.$$