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This expression is valid:

$$\left|\log(1+x) - \log(1 - x)\right| \leq \frac{\left|x-y\right|}{1+|x-y|},\ \ 0<x<1, \ 0<y<1 $$

How we can prove using Mathematica (either graphically or using $x = y$ and $x \neq y$)

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3 Answers 3

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Your claim is invalid.

Look at your expression:

Abs[Log[1 + x] - Log[1 - x]] <= Abs[x - y]/(1 + Abs[x - y])

Assume x==y, then the right hand side is zero. The left hand side is e.g. for x == y == 0.5

Abs[Log[1 + x] - Log[1 - x]]/. {x -> .5, y -> .5}
(* 1.09861 *)
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  • $\begingroup$ Sorry...It was typing mistake from my side. $\endgroup$
    – meraj
    Feb 12, 2021 at 18:02
  • $\begingroup$ It is actually,Abs[Log[1 + x] - Log[1 - y]] <= Abs[x - y]/(1 + Abs[x - y]) $\endgroup$
    – meraj
    Feb 12, 2021 at 18:02
  • $\begingroup$ @meraj please correct it in the question by editing it. $\endgroup$
    – flinty
    Jul 12, 2021 at 21:35
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Clear["Global`*"]

The inequality is not valid over the specified ranges.

ineq = Abs[Log[1 + x] - Log[1 - x]] <= Abs[x - y]/(1 + Abs[x - y]);

Plot3D[Evaluate[List @@ ineq],
 {x, 0, 1}, {y, 0, 1},
 PlotLegends -> "Expressions", ClippingStyle -> None,
 AxesLabel -> Automatic]

enter image description here

ineq /. {x -> 1/2, y -> 1/2}

(* False *)
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Just for fun (noting that in general the inequality is false):

Manipulate[
 ContourPlot[f[x, y], {x, 0, 1}, {y, 0, 1}, Contours -> {0}, 
  ContourShading -> {Red, Green}, 
  Epilog -> Text[f @@ p > 0, p, {0, -2}]], {p, {0.5, 0.5}, Locator, 
  Appearance -> "+"}, 
 Initialization :> (f[x_, y_] := 
    Abs[x - y]/(1 + Abs[x - y]) - Log[(1 + x)/(1 - x)])]

Comment: I omitted Abs in second term as in range 0<=x<1 Log expression is >=0

enter image description here

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