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I have the below function

$$ Y = (x-1)\Bigg(\frac{B - A - 1}{1 - 2x} + \frac{x \log\Bigg[\Big(\frac{1 - x}{x}\Big)\frac{\big(A(1 - 2x) + x\big)}{\big(B(1 - 2x) + x\big)}\Bigg]}{(1 - 2x)^2}\Bigg)$$

$A$ and $B$ are variables between $0$ and $1$ with the condition that $B$ is greater than $A$. I know that this function is continuously decreasing for values of $x$ between $0$ and $1$. I want to show this analytically.

I first tried to compute the derivative of $Y$ with respect to $x$ and check if it is positive or negative but Reduce returned no answer. Then I decided to check the sign of $\Big(Y(2x) - Y(x)\Big)$ but all the techniques that I know in Mathematica are failing to give me an answer (Reduce, Solve, NSolve, FullSimplify). Is there a simpler way around this that I am missing?

Sample code used:

$Assumptions = 0<A<1 && 0<B<1 && B>A 
P[x_] := (x-1)(((B-A-1)/(1-2x))+(x/(1-2x)^2) Log[((1-x)(x+A(1-2x)))/((x)(x+B(1-2x)))])
Pd[x_] := D[P[x], x]
Z = P[2 a] - P[a]
Reduce[{$Assumptions, Z <= 0 && 0<a<1}, a, Reals]

Note: The messages that I get when running these commands are either This system cannot be solved with the methods available to Reduce or Reduce was unable to solve the system with inexact coefficients or the system obtained by direct rationalization of inexact numbers present in the system. Since many of the methods used by reduce require exact input, providing reduce with an exact version of the system may help.

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  • 1
    $\begingroup$ Why didn't you provide your Mathematica code? $\endgroup$ – Ulrich Neumann Feb 5 '18 at 10:34
  • $\begingroup$ @UlrichNeumann done. $\endgroup$ – Adon Feb 5 '18 at 13:35
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Try this: Here is your derivative:

P[x_] := (x - 
     1) (((B - A - 1)/(1 - 
         2 x)) + (x/(1 - 
           2 x)^2) Log[((1 - x) (x + A (1 - 2 x)))/((x) (x + 
            B (1 - 2 x)))]);
D[P[x], x] // Simplify

(*  (1/((-1 + 
  2 x)^3))(((-1 + 2 x) ((-1 + B) x (-2 x + B (-1 + 2 x)) + 
       A^2 (-1 + 2 x) (-x + B (-1 + 2 x)) - 
       A (-2 B (1 - 2 x)^2 + B^2 (1 - 2 x)^2 + x (-3 + 4 x))))/((-x + 
       A (-1 + 2 x)) (-x + B (-1 + 2 x))) + 
  Log[((1 - x) (A + x - 2 A x))/(x (B + x - 2 B x))])   *)

and this shows it:

Manipulate[
 Plot[1/(-1 + 
     2 x)^3 (((-1 + 2 x) ((-1 + B) x (-2 x + B (-1 + 2 x)) + 
          A^2 (-1 + 2 x) (-x + B (-1 + 2 x)) - 
          A (-2 B (1 - 2 x)^2 + B^2 (1 - 2 x)^2 + 
             x (-3 + 4 x))))/((-x + A (-1 + 2 x)) (-x + 
          B (-1 + 2 x))) + 
     Log[((1 - x) (A + x - 2 A x))/(x (B + x - 2 B x))]), {x, 0, 10}],
 {A, 0, 1}, {B, 0, 1}]

with the effect:

enter image description here

Have fun!

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