Inequality regarding the logarithm

Question

I have found the inequality $$ x\leq cy+\frac{1}{\log(c)}(x-y)\log(x/y), $$ for all $x,y>0$ and $c>1$. Why is this inequality true?

Answer 1

If you have a Jstor account, the paper where this is proved is here:

https://www.jstor.org/stable/1971423?seq=3#page_scan_tab_contents

Answer 2

First of all, see that the inequality is equivalent to the following: $$ 1\leq cy+\frac{1}{\log(c)}(1-y)\log(1/y). $$ If $cy\geq 1$, the result is trivial. Let's assume otherwise $cy<1$ which implies $y<1$. After rearranging terms, we get: $$ \frac{1-cy}{1-y}\log c\leq \log(1/y). $$ Using Jensen inequality we can see that: $$ \frac{1-cy}{1-y}\log c+ \frac{cy-y}{1-y}\log 1 \leq \log( \frac{c-c^2y+cy-y}{1-y}). $$ Now it is only a matter of algebraic manipulation to see that: $$ \frac{c-c^2y+cy-y}{1-y}\leq\frac{1}y, $$ which is equivalent to: $$ (c^2-c+1)y^2-(c+1)y+1\geq 0. $$ It can be seen that $\Delta=-3(c-1)^2<0$ and $c^2-c+1>0$ and so the result follows.

Answer 3

This is not true. Take $c=2, x=2, y=1$. Then the RHS is $$2\times 1 + \ln\left(\frac{1}{2}\right)(2-1) \ln\left(\frac{2}{1}\right) < 2 $$ and not $\ge 2.$