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Let $x,y,$ and $t$ be fixed real numbers, $1<x<y$, $0<t<1$. Does the following inequality hold for some $c$

$$\frac{\log{(tx+(1-t)y)}}{\log^t{x}\log^{(1-t)}{y}}>\frac{\log{(sw+(1-s)z)}}{\log^s{w}\log^{(1-s)}{z}}$$

for $1<x<y\leq w<z$, $|z-w|\leq c|y-x|$, $c>0$, and $0<s<1$?

Intuition says this is true since $\log$ becomes ''flatter'' as its argument grows, so that the ratio on the right hand side should be closer to one when $z$ and $w$ are larger. Also how large can $c$ become?

Edit: I can more or less see how to do this with crude error bounds. I guess I am curious if there are simpler ways of seeing this.

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  • $\begingroup$ Is there no connection between $s$ and $t$? $\endgroup$
    – András Bátkai
    Commented Jul 30, 2020 at 6:40
  • $\begingroup$ @AndrásBátkai Nope. $t$ is just fixed, while $s$ free to vary. $\endgroup$
    – Josiah Park
    Commented Jul 30, 2020 at 6:41
  • $\begingroup$ Since $c$ doesn't appear in the inequality and $|z-w|\le c|y-z|$ is true for some $c>0$, it seems that condition is irrelevant. Or did you omit an upper bound on $c$? $\endgroup$
    – Brendan McKay
    Commented Jul 30, 2020 at 7:02
  • $\begingroup$ @BrendanMcKay It should hold for some fixed $c$ depending on the $x,y$ and $t$. I'm interested in larger $c$. Also, if the $z$ and $w$ become spaced out enough one can find counterexamples to the inequality. Maybe this clarifies things? $\endgroup$
    – Josiah Park
    Commented Jul 30, 2020 at 7:09
  • 2
    $\begingroup$ The second derivative of $\log\log x$ is decreasing, so just compare it with $2$ parabolas with the same leading coefficient, in which case the optimal $c$ is just $2\sqrt{t(1-t)}$ if I haven't screwed my algebra. $\endgroup$
    – fedja
    Commented Jul 30, 2020 at 13:49

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