How to prove inequality including natural log?

Question

Let $0<p<1$ and $0<q<1$ be real numbers and $0 < p + q < 1$. How can I prove the following inequality is correct or not?

$$q > \frac{\ln(\frac{p}{1-q})}{\ln(\frac{p}{1-q}) + \ln(\frac{q}{1-p})}$$

$\ln$ is natural logarithm.

Answer 1

Recall the definition of Relative Entropy. In this case:

$D(q||(1-p))=\sum\limits_{x\in\mathcal{X}} q(x)\ln\left(\frac{q(x)}{1-p(x)}\right)$.

In the binary case:

$D(q||(1-p)) = q\ln(q)-q\ln(1-p)+(1-q)\ln(1-q)-(1-q)\ln(p)\geq 0$

So,

$D(q||(1-p)) = q\ln(\frac{q}{1-p})+q\ln(\frac{p}{1-q})-\ln(\frac{p}{1-q}) \geq 0$

EDIT Use the fact that relative entropy is non-negative. The equality holds iff the two distributions are identical. The last condition actually means that two distributions 1-p and q are not identical, so the relative entropy in this case is positive.

Correction

Since $p+q<1$, then $p<1-q$ and $q<1-p$, so $\ln(\frac{p}{1-q})$ and $\ln(\frac{q}{1-p})$ are both negative, thus

$q\ln(\frac{q}{1-p})+q\ln(\frac{p}{1-q})-\ln(\frac{p}{1-q}) > 0$

$\Rightarrow q\left[ \ln(\frac{q}{1-p})+\ln(\frac{p}{1-q})\right] > \ln(\frac{p}{1-q}) $

$\Rightarrow q < \dfrac{\ln(\frac{p}{1-q})}{\ln(\frac{q}{1-p})+\ln(\frac{p}{1-q})} $

Answer 2

Since $$0\lt p\lt 1$$ $$0\lt q\lt 1$$ $$0 \lt p + q \lt 1$$ Then $$\frac{p}{1-q}\lt 1\Rightarrow\ln\left(\frac{p}{1-q}\right)\lt 0$$ $$\frac{q}{1-p}\lt 1\Rightarrow\ln\left(\frac{q}{1-p}\right)\lt 0$$ So now we have $$q\gt\frac{\ln\left(\frac{p}{1-q}\right)}{\ln\left(\frac{p}{1-q}\right) + \ln\left(\frac{q}{1-p}\right)}$$ $$\frac{q}{1-q}\ln\left(\frac{q}{1-p}\right)\lt\ln\left(\frac{p}{1-q}\right)$$ Here we can use the fact that for $0\lt x\lt 1$, we have $$\frac{x-1}{x}\lt\ln(x)\lt\frac{(x-1)(x+5)}{2(2x+1)}$$ Which yields $$\ln\left(\frac{p}{1-q}\right)\lt\frac{(p+q-1)(p-5q+5)}{2(2p-q+1)(1-q)}$$ And $$\frac{p+q-1}{1-q}\lt\frac{q}{1-q}\ln\left(\frac{q}{1-p}\right)$$ If the original inequality were true, one would expect the upper bound to be greater than the lower bound. However $$\frac{p+q-1}{1-q}\lt\frac{(p+q-1)(p-5q+5)}{2(2p-q+1)(1-q)}$$ $$2(2p-q+1)\gt p-5q+5$$ $$p+q\gt 1$$ Therefore $$\frac{q}{1-q}\ln\left(\frac{q}{1-p}\right)\not\lt\ln\left(\frac{p}{1-q}\right)$$