Estimation for $\log$ expression with two variables.

Question

Consider arbitrary $\alpha\in[0,1)$, $p\in(\frac{1}{2},1)$. Does following estimation

$$\log\Bigg(\bigg(\frac{1+\alpha}{p}\bigg)^p\bigg(\frac{1-\alpha}{1-p}\bigg)^{(1-p)}\Bigg)\le \log(2)$$

hold? Is it possible to have equality?

Answer 1

Yes, this inequality holds true. As $\log$ is concave you have

• $p \log x + (1-p) \log y \leq \log (px + (1+p)y)$ for $p \in [0,1]$

So, you get $$\log\Bigg(\bigg(\frac{1+\alpha}{p}\bigg)^p\bigg(\frac{1-\alpha}{1-p}\bigg)^{(1-p)}\Bigg) = p \log \bigg(\frac{1+\alpha}{p}\bigg) + (1-p) \log \bigg(\frac{1-\alpha}{1-p}\bigg)$$ $$\leq \log \left(p \frac{1+\alpha}{p} + (1-p)\frac{1-\alpha}{1-p} \right) = \log(2)$$

Answer 2

I want to answer when equality holds. Given an arbitrary $p\in(\frac{1}{2},1)$. As $\log$ is concave, following holds:

$$p\log(x)+(1-p)\log(y)\le\log(px+(1-p)y).$$

For $x=y$ one has equality, since

$$p\log(x)+(1-p)\log(x)=\log(x^px^{(1-p)})=\log(px+(1-p)x).$$

In the case above, this holds when

$$\frac{1+\alpha}{p}=\frac{1-\alpha}{1-p}\Leftrightarrow \alpha=2p-1.$$