For any real number $p \in (0,1)$, $\delta \in (0,1)$ and $p+\delta <1$, prove the following inequality: $$(p+\delta)\ln \frac {p+\delta}{p} +(1-p-\delta)\ln \frac {1-p-\delta}{1-p} \ge 2\delta^2$$
I have tried some appoximation on $\ln$ function such as $\ln(1+x) \le x$ and $\ln(1+x)\ge x-x^2/2$ but it doesn't work.
The hint.
We need to prove that $f(\delta)\geq0,$ where $$f(\delta)=(p+\delta)\ln \frac {p+\delta}{p} +(1-p-\delta)\ln \frac {1-p-\delta}{1-p}- 2\delta^2.$$ But by C-S $$f''(\delta)=\left(\ln\frac{p+\delta}{p}-\ln\frac{1-p-\delta}{1-p}-4\delta\right)'=$$ $$=\frac{1}{p+\delta}+\frac{1}{1-p-\delta}-4\geq\frac{(1+1)^2}{p+\delta+1-p-\delta}-4=0$$ and $$\lim\limits_{\delta\rightarrow0^+} f'(\delta)=\lim\limits_{\delta\rightarrow0^+} f(\delta)=0.$$ Can you end it now?