Estimates on logarithm

Question

Let $0<2b<a$, then $$ \left[\log\left(\frac{a}{b}\right)\right]^p\leq c\left[\frac{a-b}{b}\right]^{p-1}, $$ for some positive constant $c$. I have checked using the fact that $\log(1+\zeta)\leq \zeta$ for all $\zeta\geq 0$ for $\zeta=\frac{a-b}{b}$, we get $$ \left[\log\left(\frac{a}{b}\right)\right]^p\leq c\left[\frac{a-b}{b}\right]^{p}. $$ But we want the power to be $p-1$ on the R.H.S. Can someone kindly help me. Thanks.

Answer 1

Proof for $p \geq 2$. [See last line of the answer for the case $1<p<2$]

Let us show that the inequality holds with $c=2^{2p-1}$. What we have to show is $(log x )^{p} \leq 2^{p} (x-1)^{p-1}$ where $x =\frac a b$. Note that $x >2$. This is equivalent to showing that $y^{p} \leq 2^{p} (e^{y}-1)^{p-1}$ where $y=\log x >\log 2$. Let us prove this last inequality by considering the cases $\log 2 <y <2$ and $y \geq 2$. In the first case we have $2^{p} (e^{y}-1)^{p-1} >2^{p}>y^{p}$ because $e^{y} >2$. Since $2^{2p-1} >2^{p}$ we are done.

In the second case $2^{2p-1} (e^{y}-1)^{p-1} >2^{p} (y^{2p-2})>y^{p}$ since $e^{y} >1+\frac {y^{2}} 2$ and $2^{p} y^{p-2}>2^{p} 2^{p-2}=2^{2p-2} >1$. This finishes the proof.

EDIT: Using the inequality $e^{y} >1+\frac {y^{n}} {n!}$ with $n >\frac p {p-1}$ the above proof can be modified to get the inequality with larger constant $c$.