Let be $H:(0,1)\to\mathbb{R}$ where $H(x):=-x\ln\left(\frac{p}{x}\right)-(1-x)\ln\left(\frac{1-p}{1-x}\right)$, where $0<p<1$. We know that $H$ is infinitely many times continuously differentiable, and $$ H'(x)=\ln\left(\frac{x}{p}\right)-\ln\left(\frac{1-x}{1-p}\right) $$ and for $k\geq 2$ $$ H^{(k)}(x)=\frac{(-1)^k(k-2)!}{x^{k-1}}+\frac{(k-2)!}{(1-x)^{k-1}}. $$ Show that $H(x)\geq \frac{(x-p)^2}{2}$ for all $x\in (0,1)$. (The round parentheses mean that $0$ and $1$ are excluded.)
My approach:
I tried to show this statement by using the Taylor expansion and developed the Taylor series at $x=p$:
$\begin{align*} &H(x\mid p)=\\ & H^{0}(p\mid p) (x-p)^0+H^{1}(p\mid p) (x-p)^1+H^{2}(p\mid p) \frac{(x-p)^2}{2!}+H^{3}(p\mid p) \frac{(x-p)^3}{3!}+H^{4}(p\mid p) \frac{(x-p)^4}{4!}+\dots\\ &=0+0+\left(\frac{1}{p}+\frac{1}{1-p}\right)\frac{(x-p)^2}{2!}+H^{3}(p\mid p) \frac{(x-p)^3}{3!}+H^{4}(p\mid p) \frac{(x-p)^4}{4!}+\dots\\ \end{align*}$
As $\left(\frac{1}{p}+\frac{1}{1-p}\right)>1$ it seems like that we only have to show that the remaining summands are $\geq 0$ if we add them up. But I failed to show this...
Any ideas? Or is this the wrong way?
There is no need for references. We bring everything on the l.h.s and we get
$f(x)=\dfrac{(x-p)^{2}}{2}+xln(p/x)+(1-x)ln\dfrac{1-p}{1-x}$.
Differentiating we get $f'(x)=x-p+ln\dfrac{p}{x}-ln\dfrac{1-p}{1-x}$ and we have:
$f'(p)=0$. We also have $f''(x)=\dfrac{x-x^{2}-1}{x(1-x)}<0$ which implies that the
function $f$ is concave and therefore local maximum is a global maximum!
$f''(p)=<0$ hence $p$ is a maximizer and $f(p)$ is the maximum.
But $f(p)=0$ therefore $f(x)\leq\,0$ which is what we had to prove!