Logarithmic inequality impossible to solve?

Question

I would like to solve an inequality of this type with parameters:

Reduce[(1 + 2b)Log[w - c] >= (1 + b)Log[w - 2*c] && 0 < b < 1 && 0 < c < w 
&& w > 0, c, Reals]

But I get: This system cannot be solved with the methods available to Reduce. The same goes for Solve. Am I missing something?

Answer 1

I don't know if this is of much help but this is the best I could do:

SubtractSides[
 MultiplySides[
   ApplySides[
    Composition[PowerExpand, Power[#, 1/(1 + 2 b)] &, 
     Exp, (# /. x_*Log[y_] -> Log[y^x]) &], (1 + 2 b) Log[-c + 
        w] >= (1 + b) Log[-2 c + w]], 
   2] /. {2 (-c + w) -> u + w, (-2 c + w) -> u}, u + w]
(*0 >= -u + 2 u^((1 + b)/(1 + 2 b)) - w*)

where $u=-2 c+ w>0$

Now that $log$ has been eliminated, this inequality can possibly be further manipulated using some special mathematical techniques. I will think about that and update the post if I find any.

Answer 2

First, I believe that you need to add the condition $w>2c$ to keep $\log(w-2c)$ real. Given that $c>0$ then $(w-c)>(w-2c)>0$. Also, $(1+2b)>(1+b)$. Then you have $(1+2b)>(1+b)>0$ and $\log(w-c)>\log(w-2c)$. There are three cases

1. $\log(w-c)>\log(w-2c)>0$ and your inequality is valid for all the values provided.

2. $\log(w-c)>0$ but $\log(w-2c)<0$ and your inequality still holds for all the values provided.

3. The remaining possibly non-trivial region is $0>\log(w-c)>\log(w-2c)$. There $(w-2c)<1$. There we can see that $1\leq\frac{1+2b}{1+b}\leq1.5$ for the values of $b$ that are of interest. Now, transforming your inequality (in the case $0>\log(w-c)>\log(w-2c)$) leads to $\frac{1+2b}{1+b}\leq\frac{\log(w-2c)}{\log(w-c)}$. Now DensityPlot or RegionPlot shows the region where $\frac{\log(w-2c)}{\log(w-c)}$ is between $1$ and $1.5$ and where $w>2c$.

       DensityPlot[UnitStep[w-2*c]*UnitStep[(Log[w-2*c]/Log[w-c]-1)]*UnitStep[-(1.5-Log[w-2*c]/Log[w-c])],{w,0,4},{c,0,2},PlotRange->All,Exclusions->None,PlotPoints->100]
   

Answer 3

It is not difficult to transform your inequality to a simpler one by defining $x=c/w$ satisfying $0<x<1$, $c<w/2$ to have real solutions, and $w-2c>1$:

$$w^b\frac{(1-x)^{2 b+1}}{(1-2 x)^{b+1}}\geq 1$$

Plotting the rational function for several values of $0<b<1$, we see clearly that $x<0.5$:

GraphicsRow@(Plot[
 Evaluate@
  Table[(1 - x)^(1 + 2 b)/(1 - 2 x)^(
   1 + b), {b, Range[1/10, 9/10, 1/10]}], {x, 0, #[[1]]}, 
 PlotRange -> #[[2]], GridLines -> {None, {1}}, 
 ImageSize -> 250] & /@ {{0.1, {0.75, 1.25}}, {0.49, {-1, 9}}})

We can inspect what happens when $x\rightarrow 0$ ($c\rightarrow 0$):

Limit[(1 - x)^(1 + 2 b)/(1 - 2 x)^(1 + b), x -> 0]

1

So it seems that the inequality is always fulfilled for $x=c/w<1/2$, and $w-2c>1$.

If $x=c/w$ satisfying $0<x<1$, $c<w/2$ to have real solutions, and $w-c<1$:

$$w^b\frac{(1-x)^{2 b+1}}{(1-2 x)^{b+1}}\leq 1$$

In this case as the rational function is monotonically increasing for different values of $b$, and $w\leq 1$, therefore, I presume some range for $c$ after solving numerically the above inequality given numerical values, and satisfying the initial conditions.

Answer 4

You might find this useful, Reduce works for any rational b:

Simplify[Reduce[
   With[{b = 1/2}, (1 + 2 b) Log[w - c] >= (1 + b) Log[w - 2*c] && 
    0 < c < w/2 && w > 0], c, Reals], 
       Assumptions -> {w > 0, 0 < c < w/2, w > 0}]

The result is conditional involving root expressions.

note that we get w>1 regardless of b. (ie. the inequality is identically satisfied for 0<b<1,w>1,0<c<w/2)

here is a plot restricted to the non-trivial region 0<w<1

Plot[Evaluate[Table[
   Simplify[
     Reduce[(1 + 2 b) Log[w - c] >= (1 + b) Log[w - 2*c] && 
       0 < c < w/2 && 0 < w < 1, c, Reals], 
     Assumptions -> {w > 0, 0 < c < w/2, 0 < w < 1}][[2]], {b, 1/10, 
    1, 1/10}]], {w, 0, 1}]